Conjectures on the distribution of roots modulo a prime of a polynomial

TL;DR

This paper proposes conjectures on the distribution of polynomial roots modulo primes, suggesting a new form of equi-distribution and relating to Dirichlet's theorem for roots in arithmetic progressions.

Contribution

It introduces new conjectures on the distribution patterns of polynomial roots modulo primes, supported by numerical data and connecting to classical number theory.

Findings

Numerical data supports the conjectured distribution patterns.

Proposes a new type of equi-distribution for roots modulo primes.

Links root distribution to Dirichlet's theorem in special cases.

Abstract

For a given monic integral polynomial $f(x)$ of degree $n$, we define local roots $r_i$ of $f(x)$ for a completely decomposable prime $p$ by $r_i \in \mathbb{Z}$, $f(r_i) \equiv 0 \bmod p$ and $0 \le r_1 \le r_2 \le \dots \le r_n < p$. With numerical data, we propose a conjecture on the distribution of $(r_1/p,\dots,r_n/p)$, which is a new kind of equi-distribution, and a conjecture of the distribution of $(r_1,\dots,r_n)$ which satisfies $r_i \equiv R_i \bmod L$ for given natural numbers $L,R_1,\dots,R_n$, which is nothing but Dirichlet's theorem on an arithmetic progression in the case $n = 1$.

Equations

$$f(x)=x^n+a_{n-1}{x}^{n-1}+\cdots +a_0$$

$$\begin{array}[]{ll}LR&:=\{(l_{1},\dots,l_{n+1})\in{\mathbb{Q}}^{n+1}\mid\sum_{i=1}^{n}l_{i}\alpha_{i}=l_{n+1}\},\\[5.69054pt] LR_{0}&:=\{(l_{1},\dots,l_{n})\in\mathbb{Q}^{n}\mid\sum_{i=1}^{n}l_{i}\alpha_{i}\in\mathbb{Q}\}\end{array}$$

$$\iota :(l_1,\dots ,l_n)\mapsto (l_1,\dots ,l_n,\sum _{i=1}^n{l}_i{\alpha }_i)$$

$$1\le t:={\dim }_{\mathbb{Q}}LR={\dim }_{\mathbb{Q}}L{R}_0\le n.$$

$${\mathbf{m}}_j:=(m_{j,1},\dots ,m_{j,n})(j=1,\dots ,t),$$

$${\hat{\mathbf{m}}}_j:=\iota ({\mathbf{m}}_j)=({\mathbf{m}}_j,m_j),m_j=\sum _{i=1}^n{m}_{j,i}{\alpha }_i(j=1,\dots ,t).$$

$$\hat{\mathbf{G}}:$$

$$=\{\sigma \in S_n\mid ⟨\sigma ({\hat{\mathbf{m}}}_1),\dots ,\sigma ({\hat{\mathbf{m}}}_t){⟩}_{\mathbb{Z}}=⟨{\hat{\mathbf{m}}}_1,\dots ,{\hat{\mathbf{m}}}_t{⟩}_{\mathbb{Z}}\},$$

$$\mathbf{G}:$$

$$=\{\sigma \in S_n\mid ⟨\sigma ({\mathbf{m}}_1),\dots ,\sigma ({\mathbf{m}}_t){⟩}_{\mathbb{Z}}=⟨{\mathbf{m}}_1,\dots ,{\mathbf{m}}_t{⟩}_{\mathbb{Z}}\}.$$

$$\sigma (\mathbf{x}):=(x_{{\sigma }^{-1}(1)},\dots ,x_{{\sigma }^{-1}(n)}),\sigma ((\mathbf{x},x))$$

$$Sp{l}_X(f):=\left\{p\le X\mid f(x)\text{ is fully splitting modulo }p\right\}$$

$$\underset{X\to \infty }{\lim }\frac{\#Sp{l}_X(f)}{\#\{p\le X\}}=\frac{1}{[\mathbb{Q}(f):\mathbb{Q}]},$$

$$f(x)\equiv \prod _{i=1}^n(x-r_i)\mathrm{mod}p,$$

$$0\le r_1\le r_2\le \cdots \le r_n<p.$$

$$\sum _{i=1}^n{m}_{j,i}{r}_{\sigma (i)}=\sum _{i=1}^n{m}_{j,{\sigma }^{-1}(i)}{r}_i\equiv m_j\mathrm{mod}p(1\le {}^{\forall }j\le t),$$

$$\sum _{i=1}^n{m}_{j,i}{r}_{\sigma (i)}=\sum _{i=1}^n{m}_{j,{\sigma }^{-1}(i)}{r}_i=m_j+k_{j}p(1\le {}^{\forall }j\le t).$$

$$\sum _i{r}_i=-a_{n-1}+kp$$

$$0$$

$$<r_1\le \cdots \le r_l$$

$$<\underset{c_1}{\underbrace{p+a_1=\cdots =p+a_1}}<\cdots <\underset{c_{k_1}}{\underbrace{p+a_{k_1}=\cdots =p+a_{k_1}}}<p,$$

$$Sp{l}_X(f,\sigma ):=\{p\in Sp{l}_X(f)\mid \sum _{i=1}^n{m}_{j,i}{r}_{\sigma (i)}\equiv m_j\mathrm{mod}p(1\le {}^{\forall }j\le t)\},$$

$$Sp{l}_X(f,\sigma ,\{k_j\}):=\{p\in Sp{l}_X(f,\sigma )\mid \sum _{i=1}^n{m}_{j,i}{r}_{\sigma (i)}=m_j+k_{j}p(1\le {}^{\forall }j\le t)\},$$

$$Sp{l}_X(f,\sigma ,\{k_j\},L,\{R_i\}):=\{p\in Sp{l}_X(f,\sigma ,\{k_j\})\mid r_i\equiv R_i\mathrm{mod}L(1\le {}^{\forall }i\le n)\}$$

$$Pr(f,\sigma ):=\underset{X\to \infty }{\lim }\frac{\#Sp{l}_X(f,\sigma )}{\#Sp{l}_X(f)},$$

$$Pr(f,\sigma ,\{k_j\}):=\underset{X\to \infty }{\lim }\frac{\#Sp{l}_X(f,\sigma ,\{k_j\})}{\#Sp{l}_X(f,\sigma )},$$

$$Pr(f,\sigma ,\{k_j\},L,\{R_i\}):=\underset{X\to \infty }{\lim }\frac{\#Sp{l}_X(f,\sigma ,\{k_j\},L,\{R_i\})}{\#Sp{l}_X(f,\sigma ,\{k_j\})},$$

$$\Delta :$$

$${\hat{\mathfrak{D}}}_n:$$

$$\mathfrak{D}(f,\sigma ):$$

$$\mathfrak{D}(f,\sigma ,\{k_j\}):$$

$$\mathfrak{D}(f,\sigma )={\cup }_{\{k_j\}\in {\mathbb{Z}}^t}\mathfrak{D}(f,\sigma ,\{k_j\})\text{(a finite sum)},$$

$$\{(0,\dots ,0),(1,\dots ,1)\}\cup {\cup }_{k=1}^{n-1}\{\mathbf{x}\in \Delta \mid \sum x_i=k\}.$$

$$c=\frac{vol(\mathfrak{D}(f,\sigma ))}{Pr(f,\sigma )}$$

$$c=\sqrt{\det (({\mathbf{m}}_i,{\mathbf{m}}_j))}/\#\hat{\mathbf{G}}.$$

$${Pr}_D(f,\sigma ):$$

Full text

1. 1 Introduction and Conjectures

2. 2 Decimal expansion of a rational number

3. 3 Linear relation among roots

4. 3.1 Non-trivial linear relation

5. 3.2 Polynomial of $\deg=4$

6. 3.3 Abelian polynomial

7. 3.4 $S_{3}$

8. 4 Dependency of $Pr(f,\sigma),\mathfrak{D}(f,\sigma)$ on $\sigma$

9. 4.1 $Pr(f,\sigma)$

10. 4.2 $\mathfrak{D}(f,\sigma)$

11. 5 One-dimensional distribution

12. 5.1 Case without non-trivial linear relation among roots

13. 5.2 Case of degree $4$ with non-trivial linear relation among roots

14. 5.3 More general case

15. 6 The Weyl criterion

16. 6.1 The Stone-Weierstrass theorem

17. 6.2 The Weyl criterion

18. 6.3 Calculation of the integral

19. 6.4 Example

20. 7 Numerical data

21. 7.1 Case without non-trivial linear relation among roots

22. 7.2 Case of degree $6$ with non-trivial linear relation among roots

23. 7.3 Case of degree $4$ with non-trivial linear relation among roots

24. 7.4 Reducible polynomial

25. 7.4.1

26. 7.4.2

27. 7.4.3

28. 8 $M(f,\mu)$ and the decimal part $\left\{\frac{g(r)}{p}\right\}$ again

29. 8.1 Polynomial without non-trivial linear relations among roots

30. 8.1.1 Bernoulli polynomial

31. 8.1.2 Proof of Theorem 8.1

32. 9 Other direction

1 Introduction and Conjectures

In this monograph, a polynomial means always a monic one over the ring $\mathbb{Z}$ of integers and the letter $p$ denotes a prime number, unless specified. Let

[TABLE]

be a polynomial of degree $n$ with complex roots $\alpha_{1},\dots,\alpha_{n}$. We fix the numbering of roots once and for all, and define vector spaces $LR,LR_{0}$ over the rational number field $\mathbb{Q}$ by

[TABLE]

which depend on the numbering of roots $\alpha_{i}$. The mapping $\iota$ defined by

[TABLE]

is an isomorphism from $LR_{0}$ on $LR$ over $\mathbb{Q}$ and more strongly from $LR_{0}\cap\mathbb{Z}^{n}$ on $LR\cap\mathbb{Z}^{n+1}$ over $\mathbb{Z}$, because numbers $\alpha_{i}$ are algebraic integers. Since there is a trivial linear relation $\sum\alpha_{i}=-a_{n-1}$, the non-zero vector $(1,\dots,1,-a_{n-1})$ is always in $LR$, hence we see clearly

[TABLE]

The condition $t=n$ holds if and only if $f(x)$ is a product of linear forms in $\mathbb{Q}[x]$. The dimension $\dim_{\mathbb{Q}}\langle\alpha_{1},\dots,\alpha_{n},1\rangle_{\mathbb{Q}}$ is $n+1-t$, hence $\dim_{\mathbb{Q}}\langle\alpha_{1},\dots,\alpha_{n}\rangle_{\mathbb{Q}}=n+1-t$ or $n-t$. The equality $\dim_{\mathbb{Q}}\langle\alpha_{1},\dots,\alpha_{n},1\rangle_{\mathbb{Q}}=\dim_{\mathbb{Q}}\langle\alpha_{1},\dots,\alpha_{n}\rangle_{\mathbb{Q}}$ is equivalent to $-a_{n-1}=\frac{tr(\sum_{i}\alpha_{i})}{n}\neq 0$ if the polynomial $f(x)$ is irreducible.

We note that the vector $(0,\dots,0,1,0,\dots,0)\in\mathbb{Z}^{n}$ is in $LR_{0}$ if and only if $f(x)$ has a rational root, and $(0,\dots,0,1,0,\dots,0,-1,0,\dots,0)\in LR_{0}$ holds if and only if $\alpha_{i}-\alpha_{j}\in\mathbb{Q}$ holds for $i\not=j$, that is $f(x)$ is of the form $f(x)=g(x)g(x+r)F(x)$ for some polynomials $g(x),F(x)\in\mathbb{Q}[x]$ and an integer $r$.

We say that a polynomial $f(x)$ has a non-trivial linear relation among roots if $t>1$. We know that if there is no non-trivial linear relation, then the polynomial $f(x)$ is irreducible, and that if $f(x)$ is irreducible and $\deg f(x)=n$ is prime, then there is no non-trivial linear relation among roots (cf. §3).

We take and fix a $\mathbb{Z}$-basis of $LR_{0}\cap\mathbb{Z}^{n}$

[TABLE]

and put

[TABLE]

It is clear that the vectors $\hat{\bm{m}}_{1},\dots,\hat{\bm{m}}_{t}$ are a basis of $LR\cap\mathbb{Z}^{n+1}$ over $\mathbb{Z}$. If a vector $(l_{1},\dots,l_{n})\in\mathbb{Z}^{n}$ satisfies $\sum_{i=1}^{n}l_{i}\alpha_{i}\in\mathbb{Z}$, then we see easily $\sum_{i=1}^{n}(l_{i}/l)\alpha_{i}\in\mathbb{Z}$ for $l=\gcd(l_{1},\dots,l_{n})$, hence the matrix whose rows are $\bm{m}_{1},\dots,\bm{m}_{t}$ is primitive, that is elementary divisors of the $(t,n)$-matrix with rows $\bm{m}_{i}$ are $1$ only, hence it is complemented to an integral $(n,n)$-matrix with determinant $1$.

We introduce two groups associated with them:

[TABLE]

Here, $\langle\bm{z}_{1},\dots,\bm{z}_{k}\rangle_{\mathbb{Z}}:=\{a_{1}\bm{z}_{1}+\dots+a_{k}\bm{z}_{k}\mid a_{1},\dots,a_{k}\in\mathbb{Z}\}$ and we let, for $\bm{x}=(x_{1},\dots,x_{n})\in\mathbb{R}^{n},\,x\in\mathbb{R}$ and a permutation $\sigma\in S_{n}$

[TABLE]

Groups $\hat{\bm{G}},{\bm{G}}$ are independent of the choice of bases $\hat{\bm{m}}_{i}$ and $\hat{\bm{G}}$ is a subgroup of ${\bm{G}}$. If $f(x)$ is irreducible, then they are identical. However, they are not necessarily equal for a reducible polynomial, because $\sum_{i}l_{i}\alpha_{i},\sum_{i}l_{i}\alpha_{\sigma(i)}$ may be different even if they are rational.

If $f(x)$ has no non-trivial linear relation among roots, then $(1,\dots,1,-a_{n-1})$ is a basis of $LR\cap\mathbb{Z}^{n+1}$, hence $\hat{\bm{G}}={\bm{G}}=S_{n}$.

We fix the basis $\hat{\bm{m}}_{j}$ $(j=1,\dots,t)$ together with roots $\alpha_{i}$ once and for all.

Next, we put

[TABLE]

for a positive number $X$ and $Spl(f):=Spl_{\infty}(f)$. We know that $Spl(f)$ is an infinite set and that the density theorem due to Chebotarev holds, that is

[TABLE]

where $\mathbb{Q}(f):=\mathbb{Q}(\alpha_{1},\dots,\alpha_{n})$ is a finite Galois extension field of $\mathbb{Q}$ generated by all roots of $f(x)$ ([Se]). We note that except finitely many primes, three conditions (i) $p\in Spl(f)$, (ii) a prime ideal $(p)$ of $\mathbb{Q}$ splits completely in $\mathbb{Q}(f)$ and (iii) $f(x)$ is a product of $n$ linear forms over $\mathbb{Q}_{p}$ are equivalent.

We require the following conditions on the local roots $r_{1},\dots,r_{n}\,(\in\mathbb{Z})$ of $f(x)\equiv 0\bmod p$ for a prime $p\in Spl(f)$ :

[TABLE]

The condition (4) is nothing but the definition of $p\in Spl(f)$, and (5) determines local roots $r_{i}$ uniquely. Through this monograph, local roots $r_{i}$ are supposed to satisfy two conditions (4) and (5). The global ordering (5) to local roots is a key.

We know that for a sufficiently large prime $p\in Spl(f)$, there is at least one permutation $\sigma\in S_{n}$ dependent on $p$ such that

[TABLE]

hence there are some permutation $\sigma\in S_{n}$ and integers $k_{j}\,(j=1,\dots,t)$ dependent on the prime $p$ satisfying equations

[TABLE]

The possibility of integers $k_{j}$ is finite by the inequality $|k_{j}|p\leq\sum|m_{j,i}|p+|m_{j}|$.

Just to make sure, we give a proof of (6) here: If a prime $p\in Spl(f)$ is sufficiently large, then it is completely decomposable at $\mathbb{Q}(f)$, hence there is a prime ideal $\mathfrak{p}$ of degree $1$ over $p$. Therefore there are rational integers $r_{i}^{\prime}$ such that $\alpha_{i}\equiv r_{i}^{\prime}\bmod\mathfrak{p}$, hence $\sum_{i}m_{j,i}r_{i}^{\prime}\equiv m_{j}\bmod p$ for $j=1,\dots,t$. On the other hand, we see that $f(x)=\prod(x-\alpha_{i})\equiv\prod(x-r_{i}^{\prime})\bmod\mathfrak{p}$ and $f(x)\equiv\prod(x-r_{i})\bmod p$ imply that two sets $\{r_{i}^{\prime}\bmod p\}$ and $\{r_{i}\bmod p\}$ are equal. Hence there is a permutation $\sigma$ such that $r_{i}^{\prime}\equiv r_{\sigma(i)}\bmod p$ $(i=1,\dots,n)$.

We note that if there are infinitely many primes $p\in Spl(f)$ satisfying

[TABLE]

for an integer $k$ with $k\leq 0$ or $k\geq n$, then the polynomial $f(x)$ is a product of linear forms, because the above equation implies $0\leq r_{1}\leq\dots\leq r_{n}\leq\sum r_{i}\leq-a_{n-1}$ if $k\leq 0$, or $0>r_{n}-p\geq\dots\geq r_{1}-p\geq\sum(r_{i}-p)\geq-a_{n-1}$ if $k\geq n$, hence $f(x)\equiv\prod(x-s_{i})\bmod p$ with some integers $s_{i}$ satisfying $0\leq|s_{i}|\leq|a_{n-1}|$ for infinitely many primes $p$, which means that $f(x)$ is equal to $\prod(x-s_{i})$.

And let us give one more remark: Suppose that a polynomial $f(x)$ is of the form $\prod_{i=1}^{k_{1}}(x-a_{i})^{c_{i}}\prod_{j=1}^{k_{2}}(x-b_{j})^{d_{j}}\cdot g(x)$, where all $a_{i},b_{j}$ are integers and $a_{1}<\dots<a_{k_{1}}<0\leq b_{1}<\dots<b_{k_{2}}$ with all $c_{i},d_{j}$ being positive, and that the polynomial $g(x)$ of degree $l$ has no integer root. Then we see that for a prime $p\in Spl(f)=Spl(g)$, local roots of $f(x)$ are, with finitely many exceptions $p$

[TABLE]

where $r_{1},\dots,r_{l}$ are local roots of $g(x)$.

To state conjectures, we introduce following notations corresponding to conditions (6), (7):

[TABLE]

where $Spl_{X}(f,\sigma)$ does not depend on choice of the basis ${\bm{m}}_{i}$, but $Spl_{X}(f,\sigma,\{k_{j}\})$ does. Furthermore, for integers $L\,(>1),R_{1},\dots,R_{n}$ we put

[TABLE]

and we define densities by

[TABLE]

where for the last two, the denominators $\#Spl_{X}(f,\sigma),\#Spl_{X}(f,\sigma,\{k_{j}\})$ of the right-hand sides are supposed to tend to the infinity, and we suppose that all the limits exist. The existence of the limits is supported by computer experiment (cf. §7).

If there is no non-trivial linear relation among roots, then $Spl_{X}(f,\sigma)=Spl_{X}(f)$, i.e. $Pr(f,\sigma)=1$ and $Spl_{X}(f,\sigma,\{k_{j}\})$ is independent of the permutation $\sigma$.

Next, we introduce geometric objects

[TABLE]

where ${\mathfrak{D}}(f,\sigma)$ does not depend on the choice of basis ${\bm{m}}_{i}$, but $\mathfrak{D}(f,\sigma,\{k_{j}\})$ does. We note that $\mathfrak{D}(f,\sigma,\{k_{j}\})\subset\mathfrak{D}(f,\sigma)\subset\hat{\mathfrak{D}}_{n}\subset\Delta$ and

[TABLE]

hence ${\mathfrak{D}}(f,\sigma)$ is disconnected in general.

In case that the set $Spl(f,\sigma,\{k_{j}\})$ is an infinite set, any accumulation point $\bm{x}=(x_{1},\dots,x_{n})$ of $(r_{1}/p,\dots,r_{n}/p)\in[0,1)^{n}$ with $p\in Spl(f,\sigma,\{k_{j}\})$ satisfies $\bm{x}\in\Delta,\sum_{i}m_{j,i}x_{\sigma(i)}=k_{j}$ $(1\leq{}^{\forall}j\leq t)$, i.e. $\bm{x}\in\mathfrak{D}(f,\sigma,\{k_{j}\})$ by $\sum_{i}m_{j,i}r_{\sigma(i)}=m_{j}+k_{j}p$. The condition $x_{n}\leq 1$ can not be replaced by $x_{n}<1$. The dimension of $\hat{\mathfrak{D}}_{n}$ is $n-1$ and that of ${\mathfrak{D}}(f,\sigma)$ and ${\mathfrak{D}}(f,\sigma,\{k_{j}\})$ is less than or equal to $n-t$. In the following, the volume $vol$ of sets ${\mathfrak{D}}(f,\sigma)$ and ${\mathfrak{D}}(f,\sigma,\{k_{j}\})$ means that as an $(n-t)$-dimensional set, that is for a set $S$ in $\bm{v}+\mathbb{R}[\bm{v}_{1},\dots,\bm{v}_{n-t}](\subset\mathbb{R}^{n})$ with orthonormal vectors $\bm{v}_{1},\dots,\bm{v}_{n-t}\in\mathbb{R}^{n}$, we identify $\bm{v}+\sum y_{i}\bm{v}_{i}\in S$ with a point $(y_{1},\dots,y_{n-t})\in\mathbb{R}^{n-t}$. So, $vol({\mathfrak{D}}(f,\sigma))=0$ holds if $\dim{\mathfrak{D}}(f,\sigma)<n-t$.

In case that $f(x)$ is a product of linear forms, which is equivalent to $t=n$, $\dim{\mathfrak{D}}(f,\sigma)=0$ holds, hence the discussion about ${\mathfrak{D}}(f,\sigma)$ is meaningless.

If there is no non-trivial linear relation among roots, then ${\mathfrak{D}}(f,\sigma)$ is equal to $\hat{\mathfrak{D}}_{n}$ for every permutation $\sigma$, and $\hat{\mathfrak{D}}_{n}$ is equal to

[TABLE]

The first conjecture is

Conjecture 1**.**

For a permutation $\sigma$ with $Pr(f,\sigma)>0$, the ratio

[TABLE]

is independent of $\sigma$. If $\bm{G}=\hat{\bm{G}}$ holds, then two conditions $Pr(f,\sigma)>0$ and ${vol}({\mathfrak{D}}(f,\sigma))>0$ are equivalent.

If (i) $f(x)$ is not a product of linear forms, (ii) $\alpha_{i}-\alpha_{j}\not\in\mathbb{Q}$ if $i\neq j$, and (iii) $vol(\mathfrak{D}(f,\sigma))=cPr(f,\sigma)$ holds for every $\sigma\in S_{n}$, then we have (cf. Prop. 5.3)

[TABLE]

There is an example of a reducible polynomial $f(x)$ such that $Pr(f,\sigma)=0$ but ${vol}({\mathfrak{D}}(f,\sigma))>0$ for some permutation $\sigma$ (cf. 7.4.2), and so Expectation $1^{\prime\prime}$ in [K8], [K11] should be revised as above. The author does not know the necessary and/or sufficient condition to that two conditions $Pr(f,\sigma)>0$, ${vol}({\mathfrak{D}}(f,\sigma))>0$ are equivalent (cf. Proposition 4.5). As above, they seem to be equivalent for an irreducible polynomial $f(x)$ of $\deg f>1$. Incidentally, it is likely that $f(x)$ is a product of linear forms if and only if $vol(\mathfrak{D}(f,\sigma))=0$ holds for every permutation $\sigma$.

If $f(x)$ has no non-trivial linear relation among roots, then $Pr(f,\sigma)$ is equal to $1$ as above, and by the property ${\mathfrak{D}}(f,\sigma)=\hat{\mathfrak{D}}_{n}$ for all $\sigma\in S_{n}$ Conjecture 1 is true (cf. Proposition 4.6). If $f(x)$ is a product of linear forms, then (10) is true for $c=0$, but it is futile.

The second is

Conjecture 2**.**

Suppose that $Pr(f,\sigma)>0$ and ${vol}(\mathfrak{D}(f,\sigma))>0$ for a permutation $\sigma$; then for a set $D$ satisfying that $D=\overline{D^{\circ}}$ and $D\cap\mathfrak{D}(f,\sigma)$ is the closure of the intersection of $D^{\circ}$ and the interior of $\mathfrak{D}(f,\sigma)$, we have

[TABLE]

About the condition on $D$ above, except the measurability: In case of $m_{1}=\dots=m_{t}=0$, every point $(r_{1}/p,\dots,r_{n}/p)$ is on $\mathfrak{D}(f,\sigma)$, and so we may assume that the set $D$ is in $\mathfrak{D}(f,\sigma)$ from the beginning. This is usual and there is no problem to generalize the Weyl criterion formally, although the proof is another big problem. Suppose not, i.e. $m_{i}\neq 0$ for some $i$; then any point $(r_{1}/p,\dots,r_{n}/p)$ is not on $\mathfrak{D}(f,\sigma)$, hence for example, for $D=[0,1)^{n}\setminus\mathfrak{D}(f,\sigma)$ the left-hand side of (11) is $1$, but the right-hand side is [math]. So we must put some restriction on $D$ (cf. §6.2). Although (11) matches numerical data for the set $D=\{\bm{x}\mid x_{i}\leq a\}$, it seems to be too restrictive. The closed set whose boundaries are given by linear forms except $\sum_{i}m_{j,\sigma^{-1}(i)}x_{i}\in\mathbb{Z}$ may also be a candidate.

It is rather complicated to see whether $vol(D\cap\mathfrak{D}(f,\sigma))>0$ or not. Let us give an example when $f(x)$ has no non-trivial linear relation among roots: Let integers $k,J$ satisfy $1\leq k\leq n-1,1\leq J\leq n-k$ and

[TABLE]

Then defining $\bm{x}$ by

[TABLE]

we see that $\sum x_{j}=k$ and $\bm{x}\in D$ is an inner point of $\Delta$ if $N,\frac{M}{N}$ are sufficiently large, hence $vol(D\cap\mathfrak{D}(f,\sigma,k))>0$.

The equation (11) implies

[TABLE]

applying to

[TABLE]

The following may be better, although they are equivalent.

Conjecture 3**.**

Suppose that $Pr(f,\sigma,\{k_{j}\})>0$ and ${vol}(\mathfrak{D}(f,\sigma,\{k_{j}\}))>0$ for a permutation $\sigma$; then for a set $D$ in

[TABLE]

satisfying the same condition in Conjecture 2 we have

[TABLE]

Conjecture 2 implies Conjecture 3 as follows :

[TABLE]

and vice versa. Conjecture 2 is a kind of the equi-distribution (cf. §5). In fact, in case that there is no non-trivial linear relation among roots, (11) implies, for $1\leq{}^{\forall}i\leq n$, $0\leq{}^{\forall}a<1$

[TABLE]

where the binomial coefficient $A\choose B$ is supposed to vanish unless $0\leq B\leq A$, and $M(x):=\max(x,0)$, from which follows the one-dimensional equi-distribution

[TABLE]

In case that a polynomial $f(x)$ is quadratic and irreducible in particular, it is true by [DFI], [T]. In case that a polynomial $f(x)$ has a non-trivial linear relation among roots, see §5.3.

If $f(x)$ has no non-trivial linear relation among roots, then (12) implies with Proposition 4.6 that for $1\leq k\leq n-1$,

[TABLE]

where $E_{n}(k)$ is the volume of

[TABLE]

and it is equal to $A(n-1,k)/(n-1)!$ where Eulerian numbers $A(n,k)$ $(1\leq k\leq n)$ are defined recursively by

[TABLE]

and $\lceil x\rceil$ is the integer satisfying $x\leq\lceil x\rceil<x+1$.

Next, to state the conjecture on $Pr(f,\sigma,\{k_{j}\},L,\{R_{i}\})$ for integers $L\,(>1)$, $R_{i}$ $(i=1,\dots,n)$ as above, we introduce the following condition $(C_{1})$, which is a necessary condition to $\#Spl_{\infty}(f,\sigma,\{k_{j}\},L,\{R_{i}\})=\infty$:

$(C_{1})$ : $(k_{j},L)=(\sum_{i}m_{j,i}R_{\sigma(i)}-m_{j},L)(=d_{j}$, say) for ${}^{\forall}j=1,\dots,t$ and there is an integer $q$ which satisfies (i) $q$ is independent of $j$, (ii) $q$ is relatively prime to $L$, (iii) $\sum_{i}m_{j,i}R_{\sigma(i)}-m_{j}\equiv k_{j}\cdot q\bmod L$ $({}^{\forall}j)$ and (iv) $[[q]]=[[1]]$ on $\mathbb{Q}(f)\cap\mathbb{Q}(\zeta_{L/d_{j}})$ $({}^{\forall}j)$.

Here, $\zeta_{l}$ is a primitive $l$th root of unity, and for a subfield $F$ in $\mathbb{Q}(\zeta_{l})$ and an integer $q$ relatively prime to $l$, $[[q]]$ denotes the automorphism of $F$ induced by $\zeta_{l}\to\zeta_{l}^{q}$. The condition $[[q]]=[[1]]$ on $\mathbb{Q}(f)\cap\mathbb{Q}(\zeta_{L/d_{j}})$ is trivially satisfied if $\mathbb{Q}(f)\cap\mathbb{Q}(\zeta_{L/d_{j}})=\mathbb{Q}$. We put

[TABLE]

The last conjecture is

Conjecture 4**.**

Under the assumption $\#Spl_{\infty}(f,\sigma,\{k_{j}\})=\infty$,

[TABLE]

If the polynomial $f(x)$ has no non-trivial linear relation among roots in particular, then we see that $t=1$ and $\sum_{i}m_{j,i}R_{\sigma(i)}-m_{j}=\sum_{i}R_{i}+a_{n-1}$, hence

[TABLE]

Let us see the case of degree 1, i.e. $f(x)=x+a$ : Then we see that the permutation $\sigma$ is the identity, $t=1$, $\hat{\bm{m}}_{1}=(1,-a)$, and the local root $r_{1}$ is equal to $-a+k_{1}p$ for $k_{1}=0,1$ according to $a\leq 0$, $a>0$ if $p>|a|$. Then the condition $(C_{1})$ is $R_{1}+a\equiv 0\bmod L$, and $(R_{1}+a,L)=1$ according to $a\leq 0$, and $a>0$, respectively. Thus we have only to consider the case $k_{1}=0,a\leq 0$ or $k_{1}=1,a>0$, and neglect a finite number of primes $p$ less than or equal to $|a|$; then we see that

[TABLE]

hence $Pr(f,id)=Pr(f,id,k_{1})=1$ and

[TABLE]

by Dirichlet’s theorem, and

[TABLE]

Thus the conjecture (17) is nothing but Dirichlet’s theorem.

In case that $f(x)=x^{2}+ax+b$ is an irreducible quadratic polynomial, we have $t=1$, and if $\#Spl(f,\sigma,k)=\infty$, then $k=1$ holds, hence

[TABLE]

where $D$ is the discriminant of the quadratic field $\mathbb{Q}(f)$. Suppose that $L=2$ moreover; then $\#\mathfrak{R}(f,\sigma,1,2)=2$, and the equation $r_{1}+r_{2}=-a+p$ for a large prime $p$ implies $R_{1}+R_{2}\equiv-a+1\bmod 2$, i.e. either $\{R_{1},R_{2}\}=\{0,1\}$ in the case of $a\equiv 0\bmod 2$ or $R_{1}=R_{2}\,(=0$ or $1)$ in the case of $a\equiv 1\bmod 2$. So, the conjecture says that the density of primes $p\in Spl(f)$ with $r_{1}$ being odd is equal to that with $r_{1}$ being even.

In case of $m_{1}=\dots=m_{t}=0$, the point $(r_{1}/p,\dots,r_{n}/p)$ is on $\mathfrak{D}(f,\sigma,\{k_{j}\})$ for every $p\in Spl(f,\sigma,\{k_{j}\})$, hence the situation may be a bit simplified, although it does not mean “easy” at all: Put $\beta_{i}:=[\mathbb{Q}(f):\mathbb{Q}]\cdot\linebreak[3]\alpha_{i}-\text{tr}(\alpha_{i})$ $(i=1,\dots,n)$ and $g(x):=\prod(x-\beta_{i})(\in\mathbb{Z}[x])$. Here tr denotes the trace from $\mathbb{Q}(f)$ to $\mathbb{Q}$. Since a linear relation among roots $\sum_{i}l_{i}\alpha_{i}=l_{n+1}$ $(l_{i}\in\mathbb{Q})$ is equivalent to $\sum_{i}l_{i}\beta_{i}=0$, the vector space $LR_{0}$ is the same for polynomials $f(x),g(x)$ by $tr(\beta_{i})=0$, and the constants $m_{i}$ for the polynomial $g(x)$ are [math]. Take a prime ideal $\mathfrak{p}$ of degree $1$ of $\mathbb{Q}(f)=\mathbb{Q}(g)$ and let their local roots be $r_{i},R_{i}$: Equations $\alpha_{i}\equiv r_{\sigma(i)}\bmod\mathfrak{p},\beta_{i}\equiv R_{\mu(i)}\bmod\mathfrak{p}$ for some permutations $\sigma,\mu$ imply $[\mathbb{Q}(f):\mathbb{Q}]r_{\sigma(i)}-tr(\alpha_{i})\equiv R_{\mu(i)}\bmod p$. The relation between $\sigma,\mu$ is irregular. Even if $f(x)$ has no multiple roots, $g(x)$ may have multiple roots. Can one reduce the proof of conjectures for $f(x)$ to that of $g(x)$ nevertheless?

It may be interesting to consider another criterial numbering instead of the numbering (5). As an example, we give the order $0\leq\{qx_{1}\}\leq\dots\leq\{qx_{n}\}<1$ $(\bm{x}\in[0,1)^{n})$ for a non-zero integer $q$, where $\{x\}$ means the decimal part of $x$, i.e. $0\leq\{x\}<1,x-\{x\}\in\mathbb{Z}$. This gives the numbering of roots $\{r\mid f(r)\equiv 0\bmod p,0\leq r<p\}$ if $p$ does not divide $q$. The order (5) corresponds to $q=1$. There is no reason to consider that the similar conjectures do not exit.

We explain the above in detail after the next section. Let us brief about subsequent sections. In §$2$, we give a remark on the purely periodic decimal expansion of rational numbers: Let $p\,(>5)$ be a prime number and let the purely periodic decimal expansion of $1/p$ be

[TABLE]

where $e$ is the minimal length of periods, i.e. $e=$ the order of $10$ in $(\mathbb{Z}/p\mathbb{Z})^{\times}$. Suppose $e=ln$ for natural numbers $n\;(>1),l$; we divide the period to $n$ parts of equal length $l$, and add them. Then we have

[TABLE]

for some integer $k$ with $1\leq k\leq n-1$. Here we see that $kp$ is the sum of all roots $r_{i}$ of $x^{n}-1\equiv 0\bmod p$ with $0\leq r_{i}<p$. If we restrict the factor $n$ to a fixed prime number, the density of prime numbers $p$ corresponding to the integer $k$ is closely related to that given by (16). This is our starting observation.

In §$3$, we discuss linear relations among roots. If the degree of an irreducible polynomial is prime, then there is no non-trivial linear relation. In case that (i) the degree of an irreducible polynomial is $4$, (ii) the degree is $6$ and the polynomial is Galois, or (iv) the polynomial is abelian, we see how the structure of linear relations is with the help of representation theory of a finite group.

In §4, we see how $\mathfrak{D}(f,\sigma)$ and $Pr(f,\sigma)$ depend on $\sigma$ involving the groups $\hat{\bm{G}},{\bm{G}}$. We introduce an assistant subset

[TABLE]

where $\mathfrak{p}$ denotes a prime ideal of $\mathbb{Q}(f)$.

In §5, we show that the conjecture (11) implies the one-dimensional equi-distribution of $r_{i}/p$ $(i=1,\dots,n)$ for $p\in Spl(f)$ as a corollary of evaluation of $\lim_{X\to\infty}\#\{p\in Spl_{X}(f)\mid r_{i}/p<a\}/\#Spl_{X}(f)$ in the case that (i) $f(x)$ has no non-trivial linear relation among roots, and (ii) $f(x)$ is an irreducible polynomial of degree $4$. Without invoking forcible calculation, we show that Conjectures 1,2 imply the equi-distribution of $r_{i}/p$ for local roots $r_{i}$ of a more general polynomial in §5.3.

We discuss the Weyl criterion in §6, give numerical data in §7, and in §8 conjectures on $M(f,\mu)$ and as a question derived from it the density of the set

[TABLE]

in $Spl(f)$, where $g_{1}(x),g_{2}(x)$ are not necessarily monic polynomials over $\mathbb{Z}$ and $r_{i}$ runs over local roots of $f(x)$ at $p$ and $\{a\}$ is the decimal part of $a$, and lastly in §9 other directions.

2 Decimal expansion of a rational number

Let us begin with the example $1/7=0.\dot{1}4285\dot{7}=0.142857142857\dots$: We see

[TABLE]

The question is what numbers $1,1,3$ are. One answer is

Theorem 2.1**.**

Let $a,b$ be natural numbers satisfying $(10a,b)=1$ and $a<b$, and let the purely periodic decimal expansion of $a/b$ be

[TABLE]

where $e$ is the minimal length of periods, i.e. $e=$ the order of $10\bmod b$. Suppose $e=ln$ for natural numbers $n\,(>1),l$, and define natural numbers $B,L$ by $L=(10^{l}-1,b),\,b=BL$. We divide the period to $n$ parts of equal length $l$ , and add them; then we have

[TABLE]

where $k:=\sum_{i=0}^{n-1}s_{i}/B$ is an integer with the natural number $s_{i}$ satisfying $s_{i}\equiv 10^{li}a\bmod b$ and $0\leq s_{i}<b$. In case of $a=1,\,n=2$, $k$ is uniquely determined by

[TABLE]

Before the proof, let us give remarks. Suppose $L=1$; then $B=b$ and

[TABLE]

In case of $a=1,n=2$ moreover, $k=1$ holds, which is classical. Next, consider the case that $a=1$ and $b$ is a prime number $p$; then integers $s_{i}$ give all roots of $x^{n}-1\equiv 0\bmod p$ and $\sum s_{i}=kp$. Hence the sum of local roots $r_{i}\,(0\leq r_{i}<p)$ of $x^{n-1}+x^{n-2}+\dots+1\equiv 0\bmod p$ is $-1+kp$ (cf. (7)).

Next suppose $L>1$; letting $E$ be the order of $10\bmod L$, we decompose $l$ as $KE$ by $10^{l}\equiv 1\bmod L$. Then the equation

[TABLE]

means that $(10^{l}-1)/L$ is the $K$ times iteration of the minimal period $(10^{E}-1)/L$ of $1/L$.

We were concerned with the case that $a=1$ and $b$ is prime.

Proof of Theorem 2.1

The equation $10^{li}a\equiv s_{i}\bmod b$ implies $10^{li}a/b=\lfloor 10^{li}a/b\rfloor+s_{i}/b$, where $\lfloor x\rfloor$ is the integer satisfying $x-1<\lfloor x\rfloor\leq x$. We see

[TABLE]

hence

[TABLE]

noting $s_{n}=s_{0}$. On the other hand, by the definition of $e,n,l$, we have

[TABLE]

and then $(10^{l}-1,b)=L$ implies

[TABLE]

Hence we have $\sum_{i=0}^{n-1}s_{i}\equiv(\sum_{i=0}^{n-1}10^{li})a\equiv 0\bmod B$, that is $k$ is an integer.

Suppose that $a=1,\,n=2$; then we find $s_{0}+s_{1}=Bk$ and $s_{0}=1$, and moreover $s_{1}\equiv 10^{l}\bmod b$ implies $s_{1}\equiv 1\bmod L$, hence $Bk\equiv 2\bmod L$. They imply $Bk=1+s_{1}\leq 1+(b-1)=BL$, i.e. $1\leq k\leq L$. ∎

Let us apply Theorem 2.1 to $a=1,b=p$ for a prime number $p\,(>5)$, and let $e,l,n,B,L,k,s_{i}$ be those in the theorem. Then the condition $n>1$ implies $B=b=p$, and $s_{i}$ satisfies $s_{i}^{n}\equiv 1\bmod p$ with $s_{0}=1$. For the polynomial $f(x):=(x^{n}-1)/(x-1)=x^{n-1}+\dots+1$, $s_{i}$ $(i=1,\dots,n-1)$ are all roots of $f(x)\equiv 0\bmod p$, that is $p\in Spl(f)$ and the set $\{s_{1},\dots,s_{n-1}\}$ is equal to the set of local roots of $f(x)\bmod p$. And the integer $k$ in the theorem is equal to $(\sum_{i=0}^{n-1}s_{i}+1)/p$, which is given as $\mathfrak{s}(p)$ in [HKKN]. Let us start conversely from a natural number $n$ and the polynomial $f(x)$ above. By supposing that $n$ is a prime number, $f(x)$ has no non-trivial linear relation among roots. Under Conjecture 2, we have

[TABLE]

In [HKKN], we took as a population, instead of $Spl_{X}(f)$, the set of primes $p\,(\leq X)$ such that the order of $10$ in $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is divisible by $n$, which is a subset of $Spl_{X}(f)$, and we got the similar numerical data. So, two events $\lceil(r_{1}+\dots+r_{n-1})/p\rceil=k$ and the order of $10$ in $(\mathbb{Z}/p\mathbb{Z})^{\times}$ being divisible by $n$ seem to be independent in $Spl(f)$.

The proof of the equation (16) is quite similar.

3 Linear relation among roots

A typical example of a polynomial which has a non-trivial linear relation among roots is a reducible polynomial and a decomposable one. Let $f(x)=g(x)h(x)$ $(g(x),h(x)\in\mathbb{Z}[x])$ be a reducible polynomial; then the sum of roots of $g(x)$ is an integer, which is a non-trivial linear relation. Let $f(x)=g(h(x))$ be a decomposable polynomial, that is $1<\deg g(x),\deg h(x)<\deg f(x)$ with $g(x),h(x)\in\mathbb{Q}[x]$, and let $g(x)=\prod(x-\beta_{i})$. Then the sum of roots of $h(x)-\beta_{1}=0$ is rational, hence we get a non-trivial linear relation among roots.

Let us see how to get the linear relations when we know the generator $w$ of $\mathbb{Q}(f)$. Since $\{1,w,\dots,w^{d-1}\}$ $(d=\dim_{\mathbb{Q}}\mathbb{Q}(f))$ is a basis of $\mathbb{Q}(f)$ over $\mathbb{Q}$, we define a rational matrix $A^{(n,d)}=(\bm{a}_{1},\dots,\bm{a}_{d})$ by

[TABLE]

Then, for $(l_{1},\dots,l_{n})\in\mathbb{Q}^{n}$, the condition $\sum_{i}l_{i}\alpha_{i}\in\mathbb{Q}$ holds if and only if $(l_{1},\dots,l_{n})A=(*,0,\dots,0)$ holds. So, we see that

[TABLE]

The condition $LR_{0}=\{\bm{l}\in\mathbb{Q}^{n}\mid\bm{l}A=0\}$ holds if and only if $\bm{a}_{1}$ is contained in $\mathbb{Q}[\bm{a}_{2},\dots,\bm{a}_{d}]$. This is useful when we consider examples, but theoretically seems to be useless.

3.1 Non-trivial linear relation

Let us give a sufficient condition for a polynomial to have no non-trivial linear relation among roots.

Proposition 3.1**.**

Let $f(x)$ be an irreducible polynomial of degree $n$. If $n$ is a prime number $p$, or if the Galois group $Gal(\mathbb{Q}(f)/\mathbb{Q})$ is isomorphic to $S_{n}$ or $A_{n}$ $(n\geq 4)$ as a permutation group of roots of $f(x)$, then $f(x)$ has no non-trivial linear relation among roots.

Proof.

First, suppose that the degree of a polynomial $f(x)$ is a prime $p$, and let $\alpha_{1},\dots,\alpha_{p}$ be roots of $f(x)$, and suppose a linear relation

[TABLE]

Adding a trivial relation $\sum\alpha_{i}=tr(f)$ to it if necessary, we may assume that $\sum m_{i}\neq 0$. The Galois group $Gal(\mathbb{Q}(f)/\mathbb{Q})$ acts faithfully on the set of all roots and contains an element $\sigma$ of order $p$, since $p=[\mathbb{Q}(\alpha_{1}):\mathbb{Q}]$ divides $[\mathbb{Q}(f):\mathbb{Q}]$, hence we may assume that $(\sigma(\alpha_{1}),\dots,\sigma(\alpha_{p}))=(\alpha_{2},\dots,\alpha_{p},\alpha_{1})$ by renumbering roots. Then from the assumption (18) follows

[TABLE]

Since $\alpha_{i}$’s are not rational, the determinant of the coefficient matrix of entries $m_{i}$ vanishes, hence we have $\prod_{i=0}^{p-1}(m_{1}+\zeta^{i}m_{2}+\zeta^{2i}m_{3}+\dots+\zeta^{(p-1)i}m_{p})=0$ for a primitive $p$th root $\zeta:=\zeta_{p}$ of unity, using a formula for cyclic determinant. By the assumption $\sum m_{i}\neq 0$, we have $m_{1}+\zeta^{i}m_{2}+\zeta^{2i}m_{3}+\dots+\zeta^{(p-1)i}m_{p}=0$ for some $i$ $(0<i<p)$, which implies $m_{1}=\dots=m_{p}$, that is (18) is trivial, since $\zeta^{i}$ is still a primitive $p$th root of unity.

Next, suppose that the Galois group $Gal(\mathbb{Q}(f)/\mathbb{Q})$ is isomorphic to the symmetric group $S_{n}$. For any $1\leq i<j\leq n$, there is an automorphism $\sigma$ which induces a transposition of $\alpha_{i}$ and $\alpha_{j}$. Hence we have

[TABLE]

which implies $m_{i}(\alpha_{i}-\alpha_{j})=m_{j}(\alpha_{i}-\alpha_{j})$. By $\alpha_{i}\neq\alpha_{j}$, we have $m_{i}=m_{j}$, thus (18) is trivial.

Finally, suppose that $Gal(\mathbb{Q}(f)/\mathbb{Q})$ is the alternative group $A_{n}$ and that (18) is non-trivial. Let us show that coefficients $m_{1},\dots,m_{n}$ are mutually distinct, first. Suppose that $m_{1}=m_{2}$ ; acting an even permutation $\alpha_{1}\to\alpha_{2}\to\alpha_{3}\to\alpha_{1}(=(2,3)(1,3))$ on (18), we have

[TABLE]

which imply $(m_{1}-m_{3})(\alpha_{1}-\alpha_{3})=0$, hence $m_{2}=m_{1}=m_{3}$. Considering other $\alpha_{i}$ $(i>3)$ instead of $\alpha_{3}$, we get $m_{1}=m_{2}=\dots=m_{n}$, which contradicts the non-triviality of (18). Thus coefficients $m_{i}$ are mutually distinct. Next, considering even permutations $\{\alpha_{1}\leftrightarrow\alpha_{2},\alpha_{3}\leftrightarrow\alpha_{4}\}$,$\{\alpha_{1}\leftrightarrow\alpha_{3},\alpha_{2}\leftrightarrow\alpha_{4}\}$,$\{\alpha_{1}\leftrightarrow\alpha_{4},\alpha_{2}\leftrightarrow\alpha_{3}\}$, we get

[TABLE]

which imply

[TABLE]

hence $(\alpha_{1}-\alpha_{2})^{2}=(\alpha_{3}-\alpha_{4})^{2}$. Similarly we have $(\alpha_{1}-\alpha_{3})^{2}=(\alpha_{2}-\alpha_{4})^{2}$, hence taking the difference between them, we have $(2\alpha_{1}-\alpha_{2}-\alpha_{3})(-\alpha_{2}+\alpha_{3})=(\alpha_{2}+\alpha_{3}-2\alpha_{4})(\alpha_{3}-\alpha_{2})$, i.e. $\alpha_{1}+\alpha_{4}=\alpha_{2}+\alpha_{3}$. Similarly we find $\alpha_{1}+\alpha_{3}=\alpha_{2}+\alpha_{4}$, hence a contradiction $\alpha_{3}=\alpha_{4}$. ∎

3.2 Polynomial of $\deg=4$

In case that the degree of a polynomial is $4$, we can determine polynomials with non-trivial linear relation among roots.

Proposition 3.2**.**

Let $f=x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$ be an irreducible polynomial. If there is a non-trivial linear relation among roots of $f(x)$, then $f(x)$ is decomposable, that is $f(x)=g(h(x))$ for quadratic polynomials $g(x),h(x)$.

Proof.

Let $\alpha_{1},\dots,\alpha_{4}$ be the roots of $f(x)$. Let $G:=Gal\,(\mathbb{Q}(f)/\mathbb{Q})$ be the Galois group; then it operates faithfully on a set $\{\alpha_{1},\dots,\alpha_{4}\}$ and there is a subgroup $H$ of order $4$ in $G$, because $4=[\mathbb{Q}(\alpha_{1}):\mathbb{Q}]$ divides $|G|$. Noting that for permutations $\sigma,\mu$,

[TABLE]

we see that (i) $H$ is cyclic, (ii) $H$ is generated by $(1,2)(3,4),(1,3)(2,4)$ or (iii) $(1,2),(3,4)$ up to conjugate. Renumbering roots, we have

• (i)

$G$ is generated by the cyclic permutation $\sigma$ :

[TABLE]

• (ii)

$G$ is generated by permutations $\sigma_{1},\sigma_{2}$ :

[TABLE]

• (iii)

$G$ is generated by transpositions $\sigma_{1},\sigma_{2}$ : $\sigma_{1}(\alpha_{1})=\alpha_{2}$ and $\sigma_{2}(\alpha_{3})=\alpha_{4}$.

Suppose that $\sum_{i=1}^{4}m_{i}\alpha_{i}=m$ is non-trivial, that is ${}^{\exists}m_{i}\neq{}^{\exists}m_{j}$, and if the coefficient $a_{3}$ of the polynomial vanishes, then considering $f(x-1)$ instead of $f(x)$, we may assume that $a_{3}\neq 0$ and furthermore $\sum m_{i}\neq 0$, adding a trivial relation.

First, let us consider case (i):

By linear equations $\sum_{i}m_{i}\sigma^{j}(\alpha_{i})=m$ $(j=0,1,2,3)$, we have

[TABLE]

Since $\alpha_{i}$’s are irrational, the determinant of coefficient matrix on $m_{i}$ vanishes, i.e. $\prod_{i=0}^{3}(m_{1}+\zeta^{i}m_{2}+\zeta^{2i}m_{3}+\zeta^{3i}m_{4})=0$ for a primitive fourth root $\zeta:=\zeta_{4}$ of unity. By the assumption $\sum m_{i}\neq 0$, we have

[TABLE]

hence

• (i.1)

$m_{1}-m_{3}=m_{2}-m_{4}=0$ in the case of $i=1,3$ or

• (i.2)

$m_{1}-m_{2}+\,m_{3}-m_{4}=0$ in the case of $i=2$.

Case of (i.1), i.e. $m_{1}=m_{3},\,m_{2}=m_{4}$ :

The difference of the first row and the second row gives

[TABLE]

If $m_{1}=m_{2}$ holds, we have a contradiction $m_{1}=\dots=m_{4}$. Hence we find $\alpha_{1}+\alpha_{3}=\alpha_{2}+\alpha_{4}=-a_{3}/2$, hence $f(x)=(x^{2}+a_{3}x/2+\alpha_{1}\alpha_{3})(x^{2}+a_{3}x/2+\alpha_{2}\alpha_{4})$ is a polynomial in $x^{2}+a_{3}x/2$, that is $f(x)$ is decomposable.

Case of (i.2), hence $m_{1}+m_{3}=m_{2}+m_{4}$ :

It is easy to see that

[TABLE]

By non-triviality $(m_{1}-m_{2},\dots,m_{4}-m_{1})\neq(0,\dots,0)$, the cyclic determinant of coefficients matrix vanishes, i.e.

[TABLE]

(i.2.1) Suppose $\alpha_{1}+\zeta\alpha_{2}-\alpha_{3}-\zeta\alpha_{4}=0$, i.e. $\alpha_{1}-\alpha_{3}=-\zeta(\alpha_{2}-\alpha_{4})$. By equations $\sum m_{i}\alpha_{i}=m$ and by acting $\sigma^{2}$ on it, $m_{1}\alpha_{3}+m_{2}\alpha_{4}+m_{3}\alpha_{1}+m_{4}\alpha_{2}=m$, we have $(m_{1}-m_{3})(\alpha_{1}-\alpha_{3})+(m_{2}-m_{4})(\alpha_{2}-\alpha_{4})=0$, hence $((m_{1}-m_{3})(-\zeta)+m_{2}-m_{4})(\alpha_{2}-\alpha_{4})=0.$ Therefore we get $m_{1}=m_{3},\,m_{2}=m_{4}$ and so a contradiction $m_{1}=m_{2}=m_{3}=m_{4}$ by the assumption $m_{1}+m_{3}=m_{2}+m_{4}$.

(i.2.2) Suppose that $\alpha_{1}-\alpha_{2}+\alpha_{3}-\alpha_{4}=0$; it implies $\alpha_{1}+\alpha_{3}=\alpha_{2}+\alpha_{4}$, which implies that $f(x)$ is decomposable as above.

(i.2.3) The case of $\alpha_{1}-\zeta\alpha_{2}-\alpha_{3}+\zeta\alpha_{4}=0$ is similar to (i.2.1).

Thus we have shown that in the case of (i), $f(x)$ is decomposable.

Case (ii)

The second case gives the following equations:

[TABLE]

The sum of (19), (20) (resp. (21), (22)) gives

[TABLE]

hence if $m_{1}+m_{2}\neq m_{3}+m_{4}$ holds, then $\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}$ follows, i.e. $f(x)$ is decomposable. Hence we may assume that $m_{1}+m_{2}=m_{3}+m_{4}$. Similarly, using (19), (22) (resp. (20), (21)) , we may suppose $m_{1}+m_{4}=m_{2}+m_{3}$, and $m_{1}+m_{3}=m_{2}+m_{4}$, using (19), (21) (resp. (20), (22)). These give a contradiction $m_{1}=m_{2}=m_{3}=m_{4}$.

Case (iii).

Acting $\sigma_{1},\sigma_{2}$ on $\sum m_{i}\alpha_{i}=m$, we have

[TABLE]

which imply

[TABLE]

Since $\alpha_{i}$’s are distinct, we have

[TABLE]

Hence, the equations

[TABLE]

imply

[TABLE]

by $m_{1}\neq m_{3}$. Therefore

[TABLE]

is equal to

[TABLE]

which implies

[TABLE]

If $b_{1}\neq b_{2}$ holds, then solving them, we have $\alpha_{1}\alpha_{2},\,\alpha_{3}\alpha_{4}\in\mathbb{Q}$, which implies that $f(x)$ is reducible. Thus we have $b_{1}=b_{2}$ and then $f(x)$ is a polynomial in $x^{2}-b_{1}x$, that is decomposable. ∎

Proposition 3.3**.**

Let a polynomial $f(x)=(x^{2}+ax)^{2}+b(x^{2}+ax)+c$ $(a,b,c\in\mathbb{Z})$ be irreducible, and put

[TABLE]

Then equations $\alpha_{i,1}+\alpha_{i,2}=-a$ $(i=1,2)$ give a $\mathbb{Z}$-basis of $LR\cap\mathbb{Z}^{5}$.

Proof.

We have only to show that they are a basis of $LR$. Let

[TABLE]

be a linear relation. Using $\alpha_{i,1}+\alpha_{i,2}=-a$, we may suppose

[TABLE]

We have only to show $m_{2,2}=0$, which implies $m_{1,2}=m=0$, hence we complete the proof. Suppose that $m_{2,2}\neq 0$, and dividing $m_{2,2}$, we may assume

[TABLE]

Hence $\alpha_{1,2}$ is a root of both $g(x):=x^{2}+ax-\beta_{1}$ and $h(x):=(m_{1}x+m_{2})^{2}+~{}\linebreak a(m_{1}x+m_{2})-\beta_{2}=(m_{1}x+m_{2})^{2}+a(m_{1}x+m_{2})+b+\beta_{1}$, which are polynomials over a quadratic field $\mathbb{Q}(\beta_{1})$. Since $g(x)$ is irreducible in $\mathbb{Q}(\beta_{1})[x]$, we have $h(x)=m_{1}^{2}g(x)$, hence comparing constant terms $m_{2}^{2}+am_{2}+b+\beta_{1}=-m_{1}^{2}\beta_{1}$. Thus we find a contradiction that $\beta_{1}$ is rational. ∎

In the case of $n=6$, the polynomial $f(x)=x^{6}+2x^{5}+4x^{4}+x^{3}+2x^{2}-3x+1$ is not decomposable but has a non-trivial linear relation : For a root $\alpha$, we find

[TABLE]

and the sum of first three roots is $-1$.

In case that $n=6$ and $\mathbb{Q}(\alpha_{1})$ is a Galois extension of the rational number field $\mathbb{Q}$, we can see how linear relations among roots are in subsequent subsections. To explain it, we study the problem group-theoretically.

3.3 Abelian polynomial

In this subsection, we assume that $f(x)$ is a monic irreducible polynomial of any degree over the rational number field $\mathbb{Q}$ with a complex root $\alpha$, and $\mathbb{Q}(\alpha)/\mathbb{Q}$ is a Galois extension with Galois group $G:=Gal(\mathbb{Q}(\alpha)/\mathbb{Q})$. With the help of the representation theory, we can see how the linear relations among roots are in the case that every absolutely irreducible representation of degree $>1$ is defined over $\mathbb{Q}$.

We consider the group algebra $\mathbb{C}[G]$ and generalize the action of $G$ on the root $\alpha$ to $\mathbb{C}[G]$ as $M[\alpha]:=\sum_{g\in G}m_{g}g(\alpha)\in\mathbb{C}$ for $M=\sum_{g\in G}m_{g}g\in\mathbb{C}[G]$. In case of $M\in G$, $M[\alpha]$ is the ordinary action of the element $M\in G$ on $\alpha$, and if $M\in\mathbb{Q}[G]$, then $M[\alpha]\in\mathbb{Q}(\alpha)$ is obvious. We define the vector space $GLR$ of linear relations among roots by

[TABLE]

For the numbering $G=\{g_{1},\dots,g_{n}\}$ and $\alpha_{i}=g_{i}(\alpha)$, two conditions $\sum_{i}l_{i}g_{i}\in GLR$ and $(l_{1},\dots,l_{n})\in LR_{0}$ are equivalent.

It is known ([L]) that by making $G$ act on $\mathbb{C}[G]$ from the left, $\mathbb{C}[G]$ contains the trivial representation with multiplicity $1$ and every irreducible representation of $G$, that is by putting

[TABLE]

the $G$-stable subspace $V$ spanned by $g-\delta$ $(g\in G)$ contains all non-trivial irreducible representations and does not contain the trivial representation. The equations $\delta^{2}=\delta$, $g\delta=\delta g=\delta$ $({}^{\forall}g\in G)$, $1-\delta=-\sum_{g\not=1}(g-\delta)$, and $\delta V=V\delta=0$ are clear where $1$ is the unit of the group $G$. It is also obvious that $\mathbb{C}[G]=\mathbb{C}\delta\oplus V$ by the decomposition $\sum_{g}m_{g}g=(\sum_{g}m_{g})\delta+\sum_{g}m_{g}(g-\delta)$, which implies with $\delta\in GLR$

[TABLE]

We say that an element $M\in\mathbb{C}[G]$ is trivial if and only if $M\in\mathbb{C}\delta$. It is obvious that there is a non-trivial linear relation among roots if and only if $\dim_{\mathbb{Q}}GLR>1$.

Proposition 3.4**.**

For $M\in\mathbb{C}[G]$, $M$ is trivial if and only if $M$ annihilates $V$.

Proof.

The “only if”-part follows from $\delta V=0$. Conversely, we suppose that $MV=0$ with $M=\sum m_{g}g$, in particular $M(1-\delta)=0$ for the unit $1\in G$; then

[TABLE]

Thus we have $m_{g}=m_{1}$ for every $g\in G$, since the set of $g-\delta$ $(g\neq 1)$ is a basis of $V$. Therefore $M\in\mathbb{C}\delta$ follows. ∎

Let us note the following fundamental equation for $M\in\mathbb{Q}[G]$

[TABLE]

where $g(M[\alpha])$ is the action of $g\in G=Gal(\mathbb{Q}(\alpha)/\mathbb{Q})$ on $M[\alpha]\in\mathbb{Q}(\alpha)$ and the right-hand side is the product in the group ring $\mathbb{C}[G]$. Because, putting $M=\sum_{h\in G}m_{h}h$ with $m_{h}\in\mathbb{Q}$, we see

[TABLE]

Proposition 3.5**.**

Let $M\in\mathbb{Q}[G]$; then $M\in GLR$ holds if and only if

[TABLE]

Proof.

The equation (25) is equivalent to $(\sum_{g\in G}g(M[\alpha])g^{-1})V=0$ by (24). Suppose that $M\in GLR$; then we have $m:=M[\alpha]\in\mathbb{Q}$ by the definition and it is easy to see $\sum_{g\in G}g(M[\alpha])g^{-1}\linebreak[3]=\sum_{g\in G}g(m)g^{-1}=m|G|\delta$, and it annihilates $V$. Conversely, suppose that (25); then Proposition 3.4 implies that $\sum_{g\in G}g(M[\alpha])g^{-1}$ is trivial, that is $M[\alpha]\in\mathbb{Q}(\alpha)$ is fixed by $G=Gal(\mathbb{Q}(\alpha)/\mathbb{Q})$, i.e. $M[\alpha]$ is a rational number, hence $M\in GLR$. ∎

Let $\chi$ be an irreducible character of $G$ and denote the central idempotent in $\mathbb{C}[G]$ to $\chi$ by

[TABLE]

They satisfy

[TABLE]

where $\delta_{i,j}$ is Kronecker’s delta function. If $\chi$ is the trivial character, then $c_{\chi}=\delta$. The subspace $V$ is the direct sum of $\mathbb{C}[G]c_{\chi}$ with non-trivial irreducible characters $\chi$. If $A$ is a matrix representation corresponding to $\chi$, then $\mathbb{C}[G]c_{\chi}$ is a direct sum of minimal ideals giving the representation $A$.

If $\chi$ is an irreducible character, then the mapping $g\mapsto\sigma(\chi(g))$ is also an irreducible character for every $\sigma\in Gal(\mathbb{C}/\mathbb{Q})$. We denote it by $\sigma(\chi)$, and let $\Psi$ be a minimal set of irreducible characters such that every irreducible character is of the form $\sigma(\chi)$ for ${}^{\exists}\sigma\in Gal(\mathbb{C}/\mathbb{Q}),{}^{\exists}\chi\in\Psi$.

For an irreducible character $\chi$, we put

[TABLE]

which is an abelian extension over $\mathbb{Q}$, and denote the Galois group $Gal(\mathbb{Q}(\chi)/\mathbb{Q})$ by $Gal(\chi)$. Let $K$ be a splitting field of $G$ such that it is a Galois extension of $\mathbb{Q}$ and contains $\mathbb{Q}(\chi)$. For example, the field $\mathbb{Q}(\zeta_{e})$ with exponent $e$ of $G$ is such a field. It is easy to see that $\sigma\in Gal(\chi)$ is the identity if and only if $\sigma(\chi)=\chi$. We put

[TABLE]

which is a central idempotent satisfying $\sum_{\chi\in\Psi}C(\chi)=1$. We note that $C(1)=\delta$, hence $\sum_{\chi(\neq 1)\in\Psi}C(\chi)=1-\delta$, and (25) is equivalent to

[TABLE]

for every $\chi(\neq 1)\in\Psi$.

For an irreducible character $\chi$, we put

[TABLE]

For the trivial character $1$, we have

[TABLE]

We note that for $M\in\mathbb{Q}[G]$ with $MC(\chi)=M$, the condition $M(\sum_{g}g(\alpha)g^{-1})\linebreak[3]=0$ is equivalent to

[TABLE]

Proposition 3.6**.**

For an irreducible non-trivial character $\chi$, we have

[TABLE]

and we have a linear relation

[TABLE]

Proof.

Let $M\in LR(\chi)$; then $M=MC(\chi)$ holds by definition and the fundamental equation (24) implies $\sum_{g}g(M[\alpha])g^{-1}=0$, that is $g(M[\alpha])=0$ for every $g\in G$, hence we have $M[\alpha]=0$, i.e. (28) and hence $M\in GLR$, i.e. $M=MC(\chi)\in GLR\cdot C(\chi)$. Conversely, let $M\in GLR$; then $MC(\chi)\in\mathbb{Q}[G]$ and $(MC(\chi))C(\chi)=MC(\chi)$ are obvious. Proposition 3.5 implies $M(\sum g(\alpha)g^{-1})V\linebreak[3]=0$, which, with the condition $C(\chi)\delta=0$ implies $MC(\chi)(\sum g(\alpha)g^{-1})=0$, i.e. $MC(\chi)\in LR(\chi)$. ∎

Corollary 3.1**.**

We see that (i)

[TABLE]

(ii)* $LR_{0}$ is spanned by $(1,\dots,1),c(M)$ $(M\in V\cap GLR)$, where $c(M)$ denotes the vector $(\dots,m_{g},\dots)$ for $M=\sum_{g\in G}m_{g}g$,*

(iii)* $M[\alpha]=0$ holds for $M\in V\cap GLR$.*

Proof.

First, let us show (29). Let $M\in V\cap GLR$; then we see that $M\delta=0$ implies $M=M(1-\delta)=\sum_{\chi(\neq 1)\in\Psi}MC(\chi)$. Proposition 3.6 implies $MC(\chi)\in LR(\chi)$. Hence the left-hand side is contained in the right-hand side of (29). Conversely, suppose that $M\in LR(\chi)$ $(\chi\neq 1)$; then we see that $M=MC(\chi)\in V$ and (28) implies $M\in GLR$. Thus the right-hand side of (29) is contained in the left-hand side.

Let $\sum_{g}m_{g}g(\alpha)=m$ $(m_{g},m\in\mathbb{Q})$ be a linear relation. Putting $M:=\sum m_{g}g$, we have $M\in GLR$. Decompose as $M=\sum m_{g}g=(\sum m_{g})\delta+M^{\prime}$ with $M^{\prime}:=\sum m_{g}(g-\delta)$. Then $M^{\prime}\in V\cap GLR$ is clear. Hence there are elements $M_{\chi}\in LR(\chi)$ for $\chi(\neq 1)\in\Psi$ such that $M^{\prime}=\sum M_{\chi}$ by Corollary 3.1. Hence we have the decomposition

[TABLE]

where $m=\sum_{g}m_{g}g(\alpha)=(\sum m_{g})|G|^{-1}\sum g(\alpha)$ with $M_{\chi}[\alpha]=0$. ∎

In case that $\deg\chi=1$ or $\chi$ is defined over $\mathbb{Q}$, the structure of $LR(\chi)$ is rather easy to find as we see from now on.

Theorem 3.1**.**

Suppose that $\deg\chi=1$ and let $l$ be the order of $\chi$; then $LR(\chi)\not=\{0\}$ holds if and only if $\sum_{g\in G}\chi(g)^{d}g(\alpha)=0$ for every integer $d$ relatively prime to $l$, i.e. $c_{\sigma(\chi)}[\alpha]=0$ for ${}^{\forall}\sigma\in Gal(\chi)$, and then we have

[TABLE]

with $\dim_{\mathbb{Q}}LR(\chi)=[\mathbb{Q}(\zeta_{l}):\mathbb{Q}]$.

Proof.

We note that $\mathbb{Q}(\chi)=\mathbb{Q}(\zeta_{l})$ and the equation $hc_{\chi}=\chi(h)c_{\chi}$ holds for each element $h\in G$, hence for an element $M=MC(\chi)\in\mathbb{Q}[G]$, there is an element $m\in\mathbb{Q}(\chi)$ such that $Mc_{\chi}=mc_{\chi}$. Under this remark, let us see that the mapping $M\mapsto m$ from the set $\{M=MC(\chi)\in\mathbb{Q}[G]\}$ to $\mathbb{Q}(\chi)$ is isomorphic. Suppose that $m=0$, i.e. $Mc_{\chi}=0$; then $Mc_{\sigma(\chi)}=0$ holds, acting $\sigma$ on coefficients all together, hence $M=MC(\chi)=M\sum c_{\sigma(\chi)}=0$. Next, let $m\in\mathbb{Q}(\chi)$; then there is an element $M^{\prime}c_{\chi}=mc_{\chi}$ with $M^{\prime}\in\mathbb{Q}[G]$. It is easy to see that $M:=\sum_{\sigma}M^{\prime}c_{\sigma(\chi)}=M^{\prime}C(\chi)$ satisfies $M=MC(\chi)\in\mathbb{Q}[G]$ and $Mc_{\chi}=M^{\prime}c_{\chi}=mc_{\chi}$. Therefore, the mapping $M\mapsto m$ is isomorphic, hence $M=0$ and $m=0$ are equivalent. Note that $Mc_{\chi}=mc_{\chi}$ implies $Mc_{\sigma(\chi)}=\sigma(m)c_{\sigma(\chi)}$. Take an element $M=MC(\chi)\in\mathbb{Q}[G]$. Using the above notation, we see the following two equations

[TABLE]

and

[TABLE]

The condition $M=MC(\chi)\in LR(\chi)$ holds if and only if $M(\sum g(\alpha)g^{-1})\linebreak[3]=MC(\chi)(\sum g(\alpha)g^{-1})=0$, which means $Mc_{\sigma(\chi)}(\sum g(\alpha)g^{-1})=0$ for every $\sigma\in Gal(\chi)$. By the second equation above, it is equivalent to

[TABLE]

for every $\sigma\in Gal(\chi)$, which means either $m=0$ or $\sum_{g}\sigma(\chi(g))g^{-1}(\alpha)=0$. Thus we have $\sum_{g}\sigma(\chi(g))g^{-1}(\alpha)=0$ for every $\sigma\in Gal(\chi)$ if $LR(\chi)\neq 0$. If, conversely $\sum_{g}\sigma(\chi(g))g^{-1}(\alpha)=0$ holds for every $\sigma\in Gal(\chi)$, then $M\sum g(\alpha)g^{-1}=0$, i.e. $M\in LR(\chi)$ follows for every $M=MC(\chi)\in\mathbb{Q}[G]$. Thus the first statement of the theorem has been proven. Suppose $LR(\chi)\not=0$; then the above argument implies $LR(\chi)=\{MC(\chi)\mid M\in\mathbb{Q}[G]\}=\{\sum_{g}tr_{\mathbb{Q}(\zeta_{l})/\mathbb{Q}}(\,m\,\overline{\chi(g)}\,)g\mid m\in\mathbb{Q}(\zeta_{l})\}$ by the first equation. Since $\sum_{g}tr_{\mathbb{Q}(\zeta_{l})/\mathbb{Q}}(\,m\,\overline{\chi(g)}\,)g=0$ if and only if $m=0$, we find $\dim_{\mathbb{Q}}LR(\chi)=[\mathbb{Q}(\zeta_{l}):\mathbb{Q}]$. ∎

By this theorem, Corollary 3.1 and (30), we can understand the structure of linear relations among roots if $\mathbb{Q}(\alpha)$ is an abelian extension of $\mathbb{Q}$.

Example 3.1 Let us consider a polynomial $f(x)$ of degree $6$ with a root $\alpha$ such that $\mathbb{Q}(\alpha)$ is an abelian extension of $\mathbb{Q}$. Let $G$ be the Galois group $Gal(\mathbb{Q}(\alpha)/\mathbb{Q})$ with a generator $g_{0}$, and let $\omega$ be a third root of unity, i.e. $\omega^{2}+\omega+1=0$. We number roots by $\alpha_{i}=g_{0}^{i}(\alpha)$. Let $\chi$ be a non-trivial irreducible character of order $l$ satisfying $LR(\chi)\neq 0$. Then the relations corresponding to $LR(\chi)$ are $\sum_{g}tr_{\mathbb{Q}(\chi(g_{0}))/\mathbb{Q}}(m\overline{\chi(g)})g(\alpha)=0$ for every $m\in\mathbb{Q}(\chi(g_{0}))$.

Case of $l=2$: We have $\chi(g_{0})=-1$. Relations are $\alpha_{1}+\alpha_{3}+\alpha_{5}=\alpha_{2}+\alpha_{4}+\alpha_{6}$.

Case of $l=3$: We may assume $\chi(g_{0})=\omega$. Then relations of $LR(\chi)$ are $\alpha_{1}+\alpha_{2}+\alpha_{4}+\alpha_{5}=2\alpha_{3}+2\alpha_{6}$ for $m=1$, $\alpha_{2}+\alpha_{3}+\alpha_{5}+\alpha_{6}=2\alpha_{1}+2\alpha_{4}$ for $m=\omega$, i.e. $\alpha_{1}+\alpha_{4}=\alpha_{2}+\alpha_{5}=\alpha_{3}+\alpha_{6}$.

Case of $l=6$: We assume $\chi(g_{0})=-\omega$. Equations $\sum_{g}tr_{\mathbb{Q}(\omega)/\mathbb{Q}}(m\overline{\chi(g)})g(\alpha)\linebreak[3]=0$ for $m=1,\omega$ imply $\alpha_{1}+\alpha_{5}+2\alpha_{6}=\alpha_{2}+2\alpha_{3}+\alpha_{4}$ and $2\alpha_{1}+\alpha_{2}+\alpha_{6}=\alpha_{3}+2\alpha_{4}+\alpha_{5}$, respectively, hence the basis of $LR(\chi)$ is $\alpha_{1}-\alpha_{4}=-\alpha_{2}+\alpha_{5}=\alpha_{3}-\alpha_{6}$.

The case of $l=6$ is compatible with neither the case of $l=2$ nor $l=3$.

The following polynomials define the same field $\mathbb{Q}(\zeta_{7})$ and let $g_{0}:\zeta_{7}\mapsto\zeta_{7}^{3}$. We simply write $\zeta=\zeta_{7}$.

$f(x)=x^{6}+x^{5}+\dots+1$ has only a trivial linear relation $\sum\alpha_{i}=-1$. As stated, $G=S_{6}$, $\det((\bm{m}_{i},\bm{m}_{j}))=6$ and $Pr(f,\sigma)=1$ for every $\sigma\in S_{6}$. 2. 2.

($l=2$) $f(x)=x^{6}-4x^{5}+9x^{4}-15x^{3}+18x^{2}-16x+8$ has roots $[\alpha_{1},\dots,\alpha_{6}]=[\zeta^{5}+\zeta^{4}+1,\zeta^{5}+\zeta+1,\zeta^{3}+\zeta+1,\zeta^{3}+\zeta^{2}+1,-\zeta^{5}-\zeta^{4}-\zeta^{3}-\zeta,-\zeta^{5}-\zeta^{3}-\zeta^{2}-\zeta]$, and the basis of linear relations is $\alpha_{1}+\alpha_{3}+\alpha_{5}=\alpha_{2}+\alpha_{4}+\alpha_{6}=2$. $\#G=72$, $\det((\bm{m}_{i},\bm{m}_{j}))=9$ and calculation by computer suggests that $Pr(f,\sigma)>0$ holds if and only if $\sigma\not\in G$. 3. 3.

($l=3$) $f(x)=x^{6}-3x^{5}+9x^{4}-13x^{3}+11x^{2}-5x+1$ has roots $[\alpha_{1},\dots,\alpha_{6}]=[-\zeta^{3}-\zeta^{2}-\zeta,\zeta^{5}+\zeta^{4}+\zeta+1,\zeta^{5}+\zeta^{3}+\zeta+1,\zeta^{3}+\zeta^{2}+\zeta+1,-\zeta^{5}-\zeta^{4}-\zeta,-\zeta^{5}-\zeta^{3}-\zeta]$, and the basis of linear relations is $\sum\alpha_{i}=3$, $\alpha_{1}+\alpha_{4}=\alpha_{2}+\alpha_{5}=\alpha_{3}+\alpha_{6}(=1)$, i.e. $\alpha_{1}+\alpha_{4}=\alpha_{2}+\alpha_{5}=\alpha_{3}+\alpha_{6}=1$. $\#G=48$, $\det((\bm{m}_{i},\bm{m}_{j}))=8$, and $Pr(f,\sigma)>0$ holds if and only if $\sigma\in(4,6)G$, and then $Pr(f,\sigma)=1$. 4. 4.

($l=2,3$) $f(x)=x^{6}+14x^{4}+49x^{2}+7$ has roots $[\alpha_{1},\dots,\alpha_{6}]=[-\zeta^{5}-2\zeta^{3}-\zeta^{2}-2\zeta-1,\zeta^{5}+\zeta^{4}-\zeta^{3}-\zeta^{2},2\zeta^{5}+\zeta^{4}+\zeta^{3}+2\zeta+1,\zeta^{5}+2\zeta^{3}+\zeta^{2}+2\zeta+1,-\zeta^{5}-\zeta^{4}+\zeta^{3}+\zeta^{2},-2\zeta^{5}-\zeta^{4}-\zeta^{3}-2\zeta-1]$ and the basis of linear relations is $\sum\alpha_{i}=0$, $\alpha_{1}+\alpha_{3}+\alpha_{5}=\alpha_{2}+\alpha_{4}+\alpha_{6}(=0)$, $\alpha_{1}+\alpha_{4}=\alpha_{2}+\alpha_{5}=\alpha_{3}+\alpha_{6}(=0)$, i.e. $\alpha_{4}=-\alpha_{1},\alpha_{5}=-\alpha_{2},\alpha_{6}=-\alpha_{3},\alpha_{1}+\alpha_{3}+\alpha_{5}=0$. $\#G=12$, $\det((\bm{m}_{i},\bm{m}_{j}))=12$ and $Pr(f,\sigma)>0$ holds if and only if $\sigma\in(1,3,6)G\cup(2,5)(4,6)G$. $Pr(f,(1,3,6))=3/4,Pr(f,(2,5)(4,6))=1/4$. It is easy to see that

[TABLE]

where $C=(0,0,0,1,1,1)$ and orthnormal vectors $\delta_{1},\delta_{2}$ are given by $2\delta_{1}=(1,0,1,-1,0,-1)$, $2\sqrt{3}\delta_{2}=(-1,2,1,-1,-2,1)$ and the volume is $\sqrt{3}/8$, and

[TABLE]

where $C=(0,0,1,0,1,1)$ and orthnormal vectors $\delta_{1},\delta_{2}$ are given by $2\delta_{1}=(1,0,-1,1,0,-1)$, $2\sqrt{3}\delta_{2}=(-1,2,-1,1,-2,1)$ and the volume is $\sqrt{3}/24$, hence the ratio $\mathfrak{D}(f,\sigma)/Pr(f,\sigma)=\sqrt{3}/6=\sqrt{|\det((\bm{m}_{i},\bm{m}_{j}))|}/\#\bm{G}$ is independent of $\sigma$. 5. 5.

($l=6$) $f(x)=x^{6}-x^{5}+x^{4}-x^{3}+15x^{2}-x+29$ has roots $[\alpha_{1},\dots,\alpha_{6}]=[\zeta^{3}-\zeta^{2}-\zeta,\zeta^{5}+\zeta^{4}+2\zeta^{2}+\zeta+1,-\zeta^{5}-2\zeta^{4}-\zeta^{3}-2\zeta^{2}-\zeta-1,2\zeta^{4}+\zeta^{3}+\zeta^{2}+\zeta+1,\zeta^{5}-\zeta^{4}-\zeta,-\zeta^{5}-\zeta^{3}+\zeta]$ and the basis of linear relations is $\sum\alpha_{i}=1$ and $\alpha_{1}-\alpha_{4}=-\alpha_{2}+\alpha_{5}=\alpha_{3}-\alpha_{6}$. $\#G=12$, $\det((\bm{m}_{i},\bm{m}_{j}))=72$ and $Pr(f,\sigma)>0$ holds if and only if $\sigma$ is in one of 18 cosets of $G$ represented by $id,(2,3),(2,5),(3,4),(3,6),(4,5),(5,6),$ $(2,3)(4,5),(2,3)(5,6),(2,4)(3,5),$ $(2,5)(3,4),(3,4)(5,6),(3,5)(4,6),(4,5)(3,6),(3,5,4,6),(3,6,4,5),$ $\linebreak(2,4,3,5),(2,5,3,4).$

3.4 $S_{3}$

To see the case of $S_{3}$-extension, we need more, and from now on, we assume that the absolutely irreducible character $\chi$ is associated with a rational matrix representation $A$, hence there is a minimal ideal $\mathfrak{m}$ of a simple algebra $\mathbb{Q}[G]c_{\chi}(\subset\mathbb{Q}[G])$ with a basis $\{w_{1},\dots,w_{\chi(1)}\}$ of $\mathfrak{m}$ over $\mathbb{Q}$ such that

[TABLE]

The correspondence $g\mapsto A(g)$ is extended to an isomorphism from $\mathbb{C}[G]c_{\chi}$ to the matrix ring $M_{\chi(1)}(\mathbb{C})$ with $A(\mathbb{Q}[G]c_{\chi})=M_{\chi(1)}(\mathbb{Q})$ because of $\mathbb{Q}$ being a splitting field for the representation $A$. Through this correspondence, take $v_{i,j}\in\mathbb{Q}[G]c_{\chi}$ satisfying

[TABLE]

where $E_{j,i}$ is the matrix whose $(j,i)$-entry is $1$, otherwise [math]. Therefore the set $\{v_{i,j}\mid 1\leq i,j\leq\chi(1)\}$ is a basis of $\mathbb{Q}[G]c_{\chi}$ over $\mathbb{Q}$. The identity $C(\chi)=c_{\chi}$ is obvious by the assumption that $\chi$ is associated with a rational matrix representation $A$.

Lemma 3.1**.**

We have

[TABLE]

Proof.

Write $gv_{i,j}=\sum_{1\leq k,l\leq\chi(1)}c_{k,l}v_{k,l}$ with $c_{k,l}\in\mathbb{Q}$; then we see that $gv_{i,j}w_{m}$ is $\sum_{k,l}c_{k,l}v_{k,l}w_{m}=\sum_{k,l}c_{k,l}\delta_{k,m}w_{l}=\sum_{1\leq l\leq\chi(1)}c_{m,l}w_{l}=\sum_{1\leq k\leq\chi(1)}c_{m,k}w_{k}$. On the other hand, we find that $gv_{i,j}w_{m}\linebreak[3]=g\delta_{i,m}w_{j}=\delta_{i,m}\sum_{1\leq k\leq\chi(1)}w_{k}A(g)_{k,j}$, where $A(g)_{k,j}$ is the $(k,j)$-entry of $A(g)$. Comparing the coefficients of $w_{k}$ of two equations, we have $c_{m,k}=\delta_{i,m}A(g)_{k,j}$, i.e. $c_{k,l}=\delta_{i,k}A(g)_{l,j}$. Hence we have

[TABLE]

i.e. (31). The second equation (32) follows from

[TABLE]

∎

Lemma 3.2**.**

For $M=\sum_{1\leq k,l\leq\chi(1)}m_{k,l}v_{l,k}\in\mathbb{Q}[G]c_{\chi}$ with $m_{k,l}\in\mathbb{Q}$, we have

[TABLE]

Proof.

It is easy to see that

[TABLE]

and

[TABLE]

hence

[TABLE]

i.e. (33). ∎

Proposition 3.7**.**

Each column of the matrix $B:=\sum_{g\in G}g^{-1}(\alpha)A(g)$ is equal to a linear combination of the first column of $h(B)$ $(h\in G)$ over $\mathbb{Q}$.

Proof.

Taking elements $c_{j,g}\in\mathbb{Q}$ such that $w_{j}=\sum_{h\in G}c_{j,h}hw_{1}$, we have $gw_{j}=\sum_{h\in G}c_{j,h}ghw_{1}$, hence

[TABLE]

therefore

[TABLE]

Thus the $j$-th column of $B$ is

[TABLE]

∎

Theorem 3.2**.**

For $M=Mc_{\chi}\in\mathbb{Q}[G]$, write

[TABLE]

Then $M\in LR(\chi)$ holds if and only if

[TABLE]

Moreover it is equivalent to

[TABLE]

Proof.

For $M=Mc_{\chi}\in\mathbb{Q}[G]$, the condition $M\in LR(\chi)$ is, by definition equivalent to $M(\sum_{g}g^{-1}(\alpha)g)=0$, which is $(m_{i,j})(\sum_{g}g^{-1}(\alpha)A(g))=0^{(\chi(1))}$, i.e. (34) by Lemma 3.2. Then (34) follows from Proposition 3.7. ∎

Now let us take an irreducible polynomial $f(x)$ of degree $6$ with root $\alpha$ such that $\mathbb{Q}(\alpha)$ is a Galois extension of $\mathbb{Q}$ with Galois group isomorphic to $S_{3}$. Denote the Galois group $Gal(\mathbb{Q}(\alpha)/\mathbb{Q})$ by $G$, and let $G=\langle\sigma,\mu\rangle$ with $\sigma^{3}=\mu^{2}=1,\mu\sigma\mu=\sigma^{2}$. The characters of $G$ are

the trivial character $\chi_{1}$ with $c_{\chi_{1}}=\frac{1}{6}\sum_{g}g$, 2. 2.

the character $\chi_{2}$ of degree $1$ defined by $\chi(\sigma)=1,\chi(\mu)=-1$ with $c_{\chi_{2}}=\frac{1}{6}(\sum_{i=0}^{2}\sigma^{i}-\sum_{i=0}^{2}\sigma^{i}\mu)$, 3. 3.

the character of degree $2$ corresponding to the representation

[TABLE]

with $c_{\chi_{3}}=\frac{1}{3}(2-\sigma-\sigma^{2})\in\mathbb{Q}[G]$ and $\mathbb{Q}[G]c_{\chi_{3}}=\langle 1-\sigma,\sigma-\sigma^{2},\mu-\sigma\mu,\sigma\mu-\sigma^{2}\mu\rangle_{\mathbb{Q}}$.

For simplicity, we number roots of $f(x)$ as follows:

[TABLE]

and abbreviate as

[TABLE]

In case of $\chi_{2}$, Theorem 3.1 implies that if $LR(\chi_{2})\neq 0$, then

[TABLE]

Let us consider the case $\chi:=\chi_{3}$.

Put

[TABLE]

then it is easy to see that for $i=1,2$

[TABLE]

and

[TABLE]

Thus $v_{1,1}:=v_{1},v_{1,2}:=\sigma v_{1},v_{2,1}:=v_{2},v_{2,2}:=\sigma v_{2}$ satisfy (31) and (32), and $v_{i,j}$ are in $\mathbb{Q}[G]$ for $i,j=1,2$. These imply, for $M=c_{1}v_{1}+c_{2}\sigma v_{1}+c_{3}v_{2}+c_{4}\sigma v_{2}$

[TABLE]

Moreover we see, for $i=1,2$

[TABLE]

hence the condition $M\in LR(\chi)$ is equivalent to $C{\bf{A}}=0.$ The second column of ${\bf{A}}$ is the image of the first column by $\sigma$, hence $C{\bf{A}}=0$ is certainly equivalent to $C\times(\text{the first column of }{\bf{A}})=0$. Let us see $({\bf{A}}_{1,1},{\bf{A}}_{2,1})\neq(0,0)$. If ${\bf{A}}_{1,1}={\bf{A}}_{2,1}=0$ holds, then we get ${\bf{A}}_{1,1}+\sigma({\bf{A}}_{2,1})=2\alpha_{1}-\alpha_{2}-\alpha_{3}=0$ and $2\alpha_{2}-\alpha_{3}-\alpha_{1}=0$, acting $\sigma$. Their difference is $3(\alpha_{1}-\alpha_{2})=0$, which is a contradiction.

Let us see $LR(\chi)$ explicitly.

Proposition 3.8**.**

Continue the above and suppose that $LR(\chi)\neq\{0\}$; then there are rational numbers $a,b$ with $(a,b)\neq(0,0)$ such that

[TABLE]

and the basis of $LR(\chi)$ is given by $av_{1}+bv_{2},\sigma(av_{1}+bv_{2})$, and corresponding linear relations are spanned by $a(\alpha_{1}-\alpha_{2}-\alpha_{5}+\alpha_{6})+b(-\alpha_{2}+\alpha_{3}+\alpha_{4}-\alpha_{5})=0$ and $a(\alpha_{2}-\alpha_{3}+\alpha_{4}-\alpha_{6})+b(\alpha_{1}-\alpha_{3}+\alpha_{5}-\alpha_{6})=0$.

Proof.

Suppose $LR(\chi)\neq 0$ and take a non-zero element $M=c_{1}v_{1}+c_{2}\sigma v_{1}+c_{3}v_{2}+c_{4}\sigma v_{2}\in\mathbb{Q}[G]C(\chi)$. Then at least one of $c_{i}$ is not zero, hence there are rational numbers $a,b$ with $(a,b)\neq(0,0)$ satisfying $a{\bf{A}}_{1,1}+b{\bf{A}}_{2,1}=0$, i.e. (37). The we have \left|\begin{array}[]{rr}{\bf{A}}_{1,1}&b\\ -{\bf{A}}_{2,1}&a\end{array}\right|=\left|\begin{array}[]{rr}{\bf{A}}_{1,1}&c_{3}\\ -{\bf{A}}_{2,1}&c_{1}\end{array}\right|=\left|\begin{array}[]{rr}{\bf{A}}_{1,1}&c_{4}\\ -{\bf{A}}_{2,1}&c_{2}\end{array}\right|=0 by $C{\bf{A}}=0$. Therefore, there are rational numbers $\kappa_{1},\kappa_{2}$ such that $(c_{3},c_{1})=\kappa_{1}(b,a),(c_{4},c_{2})=\kappa_{2}(b,a)$, thus we have $M=\kappa_{1}(av_{1}+bv_{2})+\kappa_{2}(a\sigma v_{1}+b\sigma v_{2})$.

∎

Example 3.2 Let $f_{0}(x):=x^{6}+3$ with a root $\zeta$. For a primitive third root $\omega$ of unity, define the automorphisms $\sigma$, $\mu$ by $\sigma(\zeta)=\omega\zeta$ and $\mu(\zeta)=-\zeta$. Roots are $[\alpha_{1},\dots,\alpha_{6}]=[\zeta,-\zeta^{4}/2-\zeta/2,\zeta^{4}/2-\zeta/2,-\zeta,\zeta^{4}/2+\zeta/2,-\zeta^{4}/2+\zeta/2]$, which satisfy $\alpha_{1}+\alpha_{2}+\alpha_{3}=\alpha_{4}+\alpha_{5}+\alpha_{6}$, and $\alpha_{1}-\alpha_{2}-\alpha_{5}+\alpha_{6}=\alpha_{2}-\alpha_{3}-\alpha_{4}+\alpha_{5}$ and $\alpha_{2}-\alpha_{3}+\alpha_{4}-\alpha_{6}=-\alpha_{1}+\alpha_{3}-\alpha_{5}+\alpha_{6}$, that is $LR(\chi_{2})\neq 0,LR(\chi_{3})\neq 0$.

The following three polynomials give the same field $\mathbb{Q}(\zeta)$ as $f_{0}(x)$.

A polynomial $f_{1}(x):=x^{6}+6x^{5}+24x^{4}+14x^{3}+15x^{2}-12x+16$ with roots $-\zeta^{3}-\zeta^{2}-\zeta-1,\,-\zeta^{5}/2+\zeta^{4}/2-\zeta^{3}+\zeta^{2}/2+\zeta/2-1,\,\zeta^{5}/2-\zeta^{4}/2-\zeta^{3}+\zeta^{2}/2+\zeta/2-1,\,\zeta^{3}-\zeta^{2}+\zeta-1,\,-\zeta^{5}/2-\zeta^{4}/2+\zeta^{3}+\zeta^{2}/2-\zeta/2-1,\,\zeta^{5}/2+\zeta^{4}/2+\zeta^{3}+\zeta^{2}/2-\zeta/2-1$ has no non-trivial linear relation.

A polynomial $f_{2}(x):=x^{6}+6x^{5}+15x^{4}+14x^{3}+24x^{2}+24x+16$ with roots $-\zeta^{2}-\zeta-1,\,-\zeta^{5}/2+\zeta^{4}/2+\zeta^{2}/2+\zeta/2-1,\,\zeta^{5}/2-\zeta^{4}/2+\zeta^{2}/2+\zeta/2-1,\,-\zeta^{2}+\zeta-1,\,-\zeta^{5}/2-\zeta^{4}/2+\zeta^{2}/2-\zeta/2-1,\,\zeta^{5}/2+\zeta^{4}/2+\zeta^{2}/2-\zeta/2-1$ has linear relations $\sum\alpha_{i}=-6$ and $\alpha_{1}+\alpha_{2}+\alpha_{3}-\alpha_{4}-\alpha_{5}-\alpha_{6}=0$ corresponding to $\chi_{2}$ only.

A polynomial $f_{3}(x):=x^{6}+6x^{5}+24x^{4}+56x^{3}+114x^{2}+132x+67$ with roots $-\zeta^{3}-\zeta-1,\,\zeta^{4}/2-\zeta^{3}+\zeta/2-1,\,-\zeta^{4}/2-\zeta^{3}+\zeta/2-1,\,\zeta^{3}+\zeta-1,\,-\zeta^{4}/2+\zeta^{3}-\zeta/2-1,\,\zeta^{4}/2+\zeta^{3}-\zeta/2-1$ has the linear relation $\sum\alpha_{i}=-6$ and the one corresponding to $\chi_{3}$ with $a=b=1$ at (37) only.

The author does not know what the set of values $a/b$ is. Numbers $2/7,37/17,\linebreak[3]\dots$ are such examples.

**Remark ** Let $g_{1}(x)$ be a polynomial with roots $\alpha_{1},\dots,\alpha_{n}$ and $g_{2}(x)$ a polynomial with roots $\beta_{1},\dots,\beta_{m}$ and $f(x):=g_{1}(x)g_{2}(x)$. Suppose that $\mathbb{Q}(g_{1})\cap\mathbb{Q}(g_{2})=\mathbb{Q}$; then a linear relation $\sum l_{i}\alpha_{i}+\sum m_{i}\beta_{i}\in\mathbb{Q}$ with all $l_{i},m_{j}\in\mathbb{Q}$ implies obviously $\sum l_{i}\alpha_{i},\sum m_{i}\beta_{i}\in\mathbb{Q}$. However, in case of $\mathbb{Q}(g_{1})\cap\mathbb{Q}(g_{2})\not=\mathbb{Q}$ a linear relation among root of $f(x)$ is not necessarily reduced to those of $g_{1}(x),g_{2}(x)$ (cf. 7.4).

4 Dependency of $Pr(f,\sigma),\mathfrak{D}(f,\sigma)$ on $\sigma$

It is easy to see that for $\nu\in S_{n}$, the condition $\nu\in\hat{\bm{G}}$ is equivalent to that there is a unimodular matrix $A_{\nu}$ of degree $t$ such that

[TABLE]

and $\nu\in{\bm{G}}$ is equivalent to that there is a unimodular matrix $A^{\prime}_{\nu}$ of degree $t$ such that

[TABLE]

and (38) implies (39) for $A^{\prime}_{\nu}=A_{\nu}$, hence $\hat{\bm{G}}$ is a subgroup of ${\bm{G}}$. Here the unimodularity of matrices $A_{\nu},A^{\prime}_{\nu}$ is replaced by the integrality, since the elementary divisors of the matrix whose rows are $\bm{m_{i}}$ are $1$.

Proposition 4.1**.**

We have

[TABLE]

Proof.

Let $\sigma\in S_{n}$. Suppose the condition $\sum_{i}m_{j,i}\alpha_{\sigma(i)}=m_{j}$, which is clearly equivalent to $\sum_{i}m_{j,\sigma^{-1}(i)}\alpha_{i}=m_{j}$, hence there are integers $a_{i,j}$ such that

[TABLE]

which means $\sigma\in\hat{\bm{G}}$. If, conversely $\sigma$ is in $\hat{\bm{G}}$, then there are integers $a_{i,j}$ such that $\sigma(\hat{\bm{m}_{j}})=\sum_{k}a_{j,k}\hat{\bm{m}}_{k}$, i.e.

[TABLE]

Hence, we see $\sum_{i}m_{j,i}\alpha_{\sigma(i)}=\sum_{i}m_{j,\sigma^{-1}(i)}\alpha_{i}=\sum_{i,k}a_{j,k}m_{k,i}\alpha_{i}=\sum_{k}a_{j,k}m_{k}=m_{j}$. Thus we have shown the first equality. The second equality follows from that $(m_{j,1},\dots,m_{j,n},m_{n})$ $(j=1,\dots,t)$ are a basis of $LR\cap\mathbb{Z}^{n+1}$ over $\mathbb{Z}$. The statement for $G$ is proved similarly, noting that the vectors $\bm{m}_{1},\dots,\bm{m}_{t}$ are a basis of $LR_{0}\cap\mathbb{Z}^{n}=\{(l_{1},\dots,l_{n})\in\mathbb{Z}^{n}\mid\sum l_{i}\alpha_{i}\in\mathbb{Z}\}$. ∎

Corollary 4.1**.**

If the polynomial $f(x)$ is irreducible, then $\hat{\bm{G}}={\bm{G}}$ holds.

Proof.

Let a permutation $\sigma$ be in $\bm{G}$. Take $(l_{1},\dots,l_{n+1})\in LR$; then we have only to show that $l:=\sum_{i=1}^{n}l_{i}\alpha_{\sigma(i)}\in\mathbb{Q}$ is equal to $l_{n+1}$. Denoting the trace from $\mathbb{Q}(f)$ to $\mathbb{Q}$ by $tr$, we have $tr(\alpha_{1})=\dots=tr(\alpha_{n})$ and $[\mathbb{Q}(f):\mathbb{Q}]l=\sum_{i}l_{i}tr(\alpha_{\sigma(i)})=\sum_{i}l_{i}tr(\alpha_{i})=tr(\sum_{i}l_{i}\alpha_{i})=tr(l_{n+1})=[\mathbb{Q}(f):\mathbb{Q}]l_{n+1}$, hence $l=l_{n+1}$. ∎

We remark that the corollary is not necessarily true for a reducible polynomial (cf. 6.4.1, 6.4.2), that is $l=l_{n+1}$ fails in the above proof.

4.1 $Pr(f,\sigma)$

To prove the next aim, we introduce more notations:

[TABLE]

where $\mathfrak{p}$ denotes a prime ideal of $\mathbb{Q}(f)$ over $p$, and then the prime ideal $\mathfrak{p}$ is unique if $p$ is relatively prime to $\prod_{\alpha_{i}\neq\alpha_{j}}(\alpha_{i}-\alpha_{j})$, that is $p$ does not divide the integer $\prod_{\alpha_{i}\neq\alpha_{j}}(\alpha_{i}-\alpha_{j})^{2}$. $\bm{G}_{0}$ is a subgroup of $\hat{\bm{G}}$ and is isomorphic to $Gal(\mathbb{Q}(f)/\mathbb{Q})$ if $f(x)$ has no multiple root.

It is clear that $Spl(f)=\cup_{\mu\in S_{n}}M(f,\mu)$, and the condition $Gal(\mathbb{Q}(f)/\mathbb{Q})=S_{n}$ implies $M(f,\mu)=Spl(f)$. The notion $M(f,\mu)$ here is better than $M_{\mu}$ in [K11].

The set $M(f,\mu)$ depends on the numbering of roots $\alpha_{i}$’s. Let $\mathbb{Q}(f)=\mathbb{Q}(\alpha)$ for an algebraic integer $\alpha$ with monic minimal polynomial $F(x)$ and write $\alpha_{i}=g_{i}(\alpha)$ $(i=1,\dots,n)$ for polynomials $g_{i}(x)\in\mathbb{Q}[x]$. Then we see that except finitely many primes $p$

[TABLE]

where $\{x\}$ means the decimal part of $x$, i.e. $0\leq\{x\}<1$ and $x-\{x\}\in\mathbb{Z}$. We note that $\mathfrak{p}=(\alpha-r,p)$ is a prime ideal except finitely many primes $p$ and the ideal $\mathfrak{p}$ and $r\bmod p$ above are unique if the prime $p$ is sufficiently large. For an integer $r$ satisfying $F(r)\equiv 0\bmod p$ define a prime ideal $\mathfrak{p}$ by $(\alpha-r,p)$; then the equation $\alpha_{i}=g_{i}(\alpha)\equiv g_{i}(r)\bmod\mathfrak{p}$ implies that there is a permutation $\eta\in S_{n}$ such that $g_{i}(r)\equiv r_{\eta(i)}\bmod p$ $(i=1,\dots,n)$, and it is easy to see that the coset $\eta\bm{G}_{0}$ is independent of the choice of the root $r$. If an integer $r^{\prime}$ has the same properties, then for the prime ideal $(\alpha-r^{\prime},p)=\mathfrak{p}^{\prime}=\sigma(\mathfrak{p})$ with some $\sigma\in Gal(\mathbb{Q}(f)/\mathbb{Q})$, we have $\alpha_{i}=g_{i}(\alpha)\equiv g_{i}(r^{\prime})\equiv r_{\eta(i)}\bmod\mathfrak{p}^{\prime}$, hence $\sigma^{-1}(\alpha_{i})\equiv r_{\eta(i)}\equiv\alpha_{i}\bmod\mathfrak{p}$. Thus $\sigma$ is the identity on $\mathbb{Q}(f)=\mathbb{Q}(\alpha_{1},\dots,\alpha_{n})$, hence we have $\mathfrak{p}=\mathfrak{p}^{\prime}$, i.e. $r\equiv r^{\prime}\bmod p$.

Problem Let $f(x)$ be an irreducible polynomial and let $g_{l}(x),h_{l}(x)$ and $G_{l}(x_{1},\dots,x_{m}),H_{l}(x_{1},\dots,x_{m})$ be polynomials in $\mathbb{Q}[x],\mathbb{Q}[x_{1},\dots,x_{m}]$ $(l=1,\dots,\linebreak[3]m)$, respectively. What is the density of the following set in $Spl(f)$ ?

[TABLE] 2. 2.

[TABLE]

Here the numbers $g_{l}(r),h_{l}(r),G_{l}(r_{1},\dots,r_{n})$ and $H_{l}(r_{1},\dots,r_{n})$ are supposed to be integers which correspond to elements in $\mathbb{Z}/p\mathbb{Z}$ for primes $p$ not dividing denominators of $g_{l}(x),h_{l}(x),G_{l}(x_{1},\dots,x_{n})$ and $H_{l}(x_{1},\dots,x_{n})$, respectively.

The computer experiment suggests that in case of $\deg f=3$ the second density above depends on $f(x)$ if it is Galois, but does not depend otherwise.

These are discussed in §8 again.

Lemma 4.1**.**

Suppose $\#M(f,\sigma)=\infty$; then for $\eta\in S_{n}$ the following three conditions are equivalent : (i) $M(f,\eta)=M(f,\sigma)$, (ii) $\sigma^{-1}\eta\in\bm{G}_{0}$, and (iii) $\#(M(f,\eta)\cap M(f,\sigma))=\infty$.

Proof.

Suppose the condition (iii), i.e. $\alpha_{i}\equiv r_{\eta(i)}\bmod\mathfrak{p}$ and $\alpha_{i}\equiv r_{\sigma(i)}\bmod\tilde{\mu}(\mathfrak{p})$ for some $\tilde{\mu}\in Gal(\mathbb{Q}(f)/\mathbb{Q})$ for a sufficiently large $p\in Spl(f)$. They imply $\tilde{\mu}(\alpha_{i})\equiv\tilde{\mu}(r_{\eta(i)})\equiv r_{\eta(i)}\bmod\tilde{\mu}(\mathfrak{p})$ and $\alpha_{\sigma^{-1}\eta(i)}\equiv r_{\eta(i)}\bmod\tilde{\mu}(\mathfrak{p})$, hence $\tilde{\mu}(\alpha_{i})\equiv\alpha_{\sigma^{-1}\eta(i)}\bmod\tilde{\mu}(\mathfrak{p})$, which imply $\tilde{\mu}(\alpha_{i})=\alpha_{\sigma^{-1}\eta(i)}$. Thus the definition implies $\sigma^{-1}\eta\in\bm{G}_{0}$, hence (ii). Suppose (ii); then $\sigma=\eta\mu^{-1}$ for some $\mu\in\bm{G}_{0}$ with $\tilde{\mu}(\alpha_{i})=\alpha_{\mu(i)}$ for some $\tilde{\mu}\in Gal(\mathbb{Q}(f)/\mathbb{Q})$. For a prime $p\in Spl(f)$, we see that $p\in M(f,\sigma)\leftrightarrow\alpha_{i}\equiv r_{\sigma(i)}\bmod\mathfrak{p}\leftrightarrow\alpha_{i}\equiv r_{\eta\mu^{-1}(i)}\bmod\mathfrak{p}\leftrightarrow\alpha_{\mu(i)}\equiv r_{\eta(i)}\bmod\mathfrak{p}\leftrightarrow\tilde{\mu}(\alpha_{i})\equiv r_{\eta(i)}\bmod\mathfrak{p}\leftrightarrow\alpha_{i}\equiv r_{\eta(i)}\bmod\tilde{\mu}^{-1}(\mathfrak{p})\leftrightarrow p\in M(f,\eta)$, hence we get (i). The condition (i) implies obviously (iii). ∎

The following is fundamental.

Proposition 4.2**.**

We have

[TABLE]

where $\mu$ runs over the set of permutations satisfying $\mu\in\sigma\hat{\bm{G}}$ with $\#M(f,{\mu})=\infty$, and $T_{\sigma}$ is a finite set. Suppose $\#Spl(f,\sigma)=\infty$; then for $\nu\in S_{n}$ we see that $Spl(f,\sigma)\cap Spl(f,\nu)$ is an infinite set if and only if $\sigma\hat{\bm{G}}=\nu\hat{\bm{G}}$, and then the difference between $Spl(f,\sigma)$ and $Spl(f,\nu)$ is a finite set.

Proof.

By the identity $Spl(f,\sigma)=\cup_{\mu\in S_{n}}(Spl(f,\sigma)\cap M(f,\mu))$, we have only to show that $\#(Spl(f,\sigma)\cap M(f,\mu))=\infty$ holds if and only if both $\mu\hat{G}=\sigma\hat{G}$ and $\#M(f,{\mu})=\infty$ hold, and then $M(f,\mu)\subset Spl(f,\sigma)$. Suppose that $\#(Spl(f,\sigma)\cap M(f,{\mu}))=\infty$. The property $\#M(f,\mu)=\infty$ is clear. For $p\in Spl(f,\sigma)\cap M(f,{\mu})$, we have

[TABLE]

which implies $\sum_{i}m_{j,i}\alpha_{\mu^{-1}\sigma(i)}\equiv m_{j}\bmod\mathfrak{p}$ for infinitely many primes in $p\in Spl(f,\sigma)\cap M(f,{\mu})$, which implies $\sum_{i}m_{j,i}\alpha_{\mu^{-1}\sigma(i)}=m_{j}$, hence we have $\mu^{-1}\sigma\in\hat{\bm{G}}$, i.e. $\mu\hat{\bm{G}}=\sigma\hat{\bm{G}}$.

Conversely, suppose that $\mu\hat{\bm{G}}=\sigma\hat{\bm{G}}$ and $\#M(f,{\mu})=\infty$ hold; then we have the equation $\sum_{i}m_{j,i}\alpha_{\mu^{-1}\sigma(i)}\linebreak[3]=m_{j}$. Hence, for $p\in M(f,\mu)$, we see $\sum_{i}m_{j,i}r_{\sigma(i)}\equiv m_{j}\bmod\mathfrak{p}$, that is $p\in Spl(f,\sigma)$ and so $M(f,\mu)\subset Spl(f,\sigma)$, thus $\#(Spl(f,\sigma)\cap M(f,\mu))=\infty$.

Hence the condition $\#(Spl(f,\sigma)\cap M(f,{\mu}))=\infty$ is equivalent to $\#M(f,{\mu})=\infty$ and $\mu\hat{\bm{G}}=\sigma\hat{\bm{G}}$. And then, we have $M(f,\mu)\subset Spl(f,\sigma)$ as above. Therefore (40) has been proved. If $Spl(f,\sigma)\cap Spl(f,\nu)$ is an infinite set, then there are permutations $\mu,\mu^{\prime}$ such that $\sigma\hat{\bm{G}}=\mu\hat{\bm{G}},\nu\hat{\bm{G}}=\mu^{\prime}\hat{\bm{G}}$ and $\#(M(f,\mu)\cap M(f,\mu^{\prime}))=\infty$, hence $\mu\bm{G}_{0}={\mu^{\prime}}\bm{G}_{0}$ by Lemma 4.1 and then $\sigma\hat{\bm{G}}=\nu\hat{\bm{G}}$. Hence the difference between $Spl(f,\sigma)$ and $Spl(f,\nu)$ is that between $T_{\sigma}$ and $T_{\nu}$. If, conversely $\sigma\hat{\bm{G}}=\nu\hat{\bm{G}}$, then from (40) follows that the difference between $Spl(f,\sigma)$ and $Spl(f,\nu)$ is finite. This completes the proof. ∎

It is likely that the density of $M(f,\mu)$ in $Spl(f,\sigma)\,(=Spl(f,\mu))$ at (40) equal to $1/[\hat{\bm{G}}:\bm{G}_{0}]$.

Corollary 4.2**.**

If a prime $p\in Spl(f)$ is sufficiently large, then there are exactly $\#\hat{\bm{G}}$ permutations $\sigma$ satisfying $p\in Spl(f,\sigma)$, i.e. (6).

Proof.

Let $p\in Spl(f)$ be large so that $p\not\in T_{\sigma}$ and $p\not\in M(f,\sigma)\cap M(f,\mu)$ if $\#(M(f,\sigma)\cap M(f,\mu))<\infty$ for $\sigma,\mu\in S_{n}$. If $p\in Spl(f)$ is contained in $Spl(f,\sigma)$, then $p\in Spl(f,\sigma\nu)$ $({}^{\forall}\nu\in\hat{\bm{G}})$ is clear by Proposition 4.2. If $p\in Spl(f)$ is contained in $Spl(f,\sigma)\cap Spl(f,\mu)$, hence in $M(f,{\sigma^{\prime}})\cap M(f,{\mu^{\prime}})$ for ${}^{\exists}\sigma^{\prime}\in\sigma\hat{\bm{G}},{}^{\exists}\mu^{\prime}\in\mu\hat{\bm{G}}$, then $M(f,{\sigma^{\prime}})\cap M(f,{\mu^{\prime}})$ is an infinite set, hence we see ${\sigma^{\prime}}^{-1}\mu^{\prime}\in\bm{G}_{0}$ and then $\sigma^{-1}\mu\in\hat{\bm{G}}$. Therefore $p\in Spl(f,\sigma)$ holds for a permutation $\sigma$ in the only one coset by $\hat{\bm{G}}$. ∎

Corollary 4.3**.**

We have

[TABLE]

Proof.

By Corollary 4.2, we see

[TABLE]

∎

Corollary 4.4**.**

Let $\sigma\in S_{n}$. If a sufficiently large prime $p\in Spl(f)$ satisfies $\sum_{i}m_{j,i}\alpha_{\sigma(i)}\equiv m_{j}\bmod\mathfrak{p}$ $(j=1,\dots,t)$ for some prime ideal $\mathfrak{p}$ of $\mathbb{Q}(f)$ lying over $p$, then $\sum_{i}m_{j,i}\alpha_{\sigma(i)}=m_{j}$ $(j=1,\dots,t)$ holds.

Proof.

Take a permutation $\kappa\in S_{n}$ such that $\alpha_{\sigma(i)}\equiv r_{\kappa(i)}\bmod\mathfrak{p}$ $(i=1,\dots,n)$. Then we see $\sum_{i}m_{j,i}r_{\kappa(i)}\equiv m_{j}\bmod p$ $(j=1,\dots,t)$, i.e. $p\in Spl(f,\kappa)$, hence $p\in M(f,{\kappa\eta})$ for some $\eta\in\hat{G}$ with $\#M(f,{\kappa\eta})=\infty$. Therefore we have $\alpha_{i}\equiv r_{\kappa\eta(i)}\bmod\mathfrak{p}^{\prime}$ $(i=1,\dots,n)$ for some prime ideal $\mathfrak{p}^{\prime}$ lying over $p$. For an automorphism $\iota\in Gal(\mathbb{Q}(f)/\mathbb{Q})$ satisfying $\iota(\mathfrak{p}^{\prime})=\mathfrak{p}$, we have $\iota(\alpha_{i})\equiv r_{\kappa\eta(i)}\bmod\mathfrak{p}$, hence $\iota(\alpha_{\eta^{-1}(i)})\equiv r_{\kappa(i)}\equiv\alpha_{\sigma(i)}\bmod\mathfrak{p}$, hence $\iota(\alpha_{\eta^{-1}(i)})=\alpha_{\sigma(i)}$. Since $\hat{G}$ is a group, we see $\eta^{-1}\in\hat{G}$, hence $\sum_{i}m_{j,i}\alpha_{\eta^{-1}(i)}=m_{j}$ $(j=1,\dots,t)$, which implies $\sum_{i}m_{j,i}\alpha_{\sigma(i)}=m_{j}$ $(j=1,\dots,t)$.

∎

By keeping notations in the proof above, the proof shows : Starting from $\sum_{i}m_{j,i}r_{\kappa(i)}\equiv m_{j}\bmod p$ and taking $\sigma\in S_{n}$ satisfying $r_{\kappa(i)}\equiv\alpha_{\sigma(i)}\bmod\mathfrak{p}$, we have $\sum_{i}m_{j,i}\alpha_{\sigma(i)}=m_{j}$. Incidentally the completion $\mathbb{Q}(f)_{\mathfrak{p}}$ is isomorphic to $\mathbb{Q}_{p}$ and let $\hat{r}_{i}$ be roots of $f(x)=0$ in $\mathbb{Q}_{p}$ satisfying $\hat{r}_{i}\equiv r_{i}\bmod p\mathbb{Z}_{p}$. Making use of the isomorphism from $\mathbb{Q}(f)$ into $\mathbb{Q}(f)_{\mathfrak{p}}\cong\mathbb{Q}_{p}$ and $\alpha_{i}\equiv r_{\kappa\sigma^{-1}}(i)\bmod\mathfrak{p}$, we see that the root $\alpha_{i}$ in $\mathbb{Q}(f)$ is mapped to the root $\hat{r}_{\kappa\sigma^{-1}(i)}$. Thus, we see that for $l_{1},\dots,l_{n+1}\in\mathbb{Q}$, the identity $\sum_{i=1}^{n}l_{i}\alpha_{i}=l_{n+1}$ is equivalent to $\sum_{i=1}^{n}l_{i}\hat{r}_{\kappa\sigma^{-1}(i)}=l_{n+1}$, and combining the above, $\sum_{i}m_{j,i}r_{\kappa(i)}\equiv m_{j}\bmod p$ implies $\sum_{i}m_{j,\sigma^{-1}(i)}\hat{r}_{\kappa\sigma^{-1}(i)}=m_{j}$, i.e. $\sum_{i}m_{j,i}\hat{r}_{\kappa(i)}=m_{j}$ in $\mathbb{Q}_{p}$ .

Proposition 4.3**.**

Let $\sigma,\nu$ be permutations, and suppose that $\nu\in\hat{\bm{G}}$. Then we have, neglecting a finite set of primes

[TABLE]

where ${}^{t}(k_{1}^{\prime},\dots,k_{t}^{\prime}):=A_{\nu}\cdot{}^{t}(k_{1},\dots,k_{t})$ for the integral matrix $A_{\nu}=(a_{ij})\in GL_{t}(\mathbb{Z})$ given at (38). In particular, we have

[TABLE]

Proof.

The first equation follows from Proposition 4.2. Let $p$ be a prime in $Spl(f,\sigma,\{k_{j}\})$; then we see $\sum_{i}m_{j,i}r_{\sigma(i)}=m_{j}+k_{j}p$ and so $\sum_{j}a_{l,j}\sum_{i}m_{j,i}r_{\sigma(i)}\linebreak[3]=\sum_{j}a_{l,j}m_{j}+\sum_{j}a_{l,j}k_{j}p$, that is $\sum_{i}m_{l,\nu^{-1}(i)}r_{\sigma(i)}\linebreak[3]=m_{l}+k_{l}^{\prime}p$, which implies $p\in Spl(f,\sigma\nu,\{k_{j}^{\prime}\})$, that is $Spl(f,\sigma,\{k_{j}\})$ is included in $Spl(f,\sigma\nu,\{k^{\prime}_{j}\})$. Since $A_{\nu}^{-1}$ is also integral, we have the converse inclusion $Spl(f,\sigma\nu,\{k_{j}^{\prime}\})\subset Spl(f,\sigma,\{k_{j}\})$ similarly, hence (42), (43). ∎

4.2 $\mathfrak{D}(f,\sigma)$

By definition, we see

[TABLE]

Another aim in this section is

Proposition 4.4**.**

Suppose that $vol({\mathfrak{D}}(f,\sigma))>0$; then we have following equivalences.

[TABLE]

In particular, ${\mathfrak{D}}(f,\sigma)={\mathfrak{D}}(f,\sigma\nu)$ holds if and only if $\nu\in{\bm{G}}$.

Proof.

For an integral vector $\bm{k}=(k_{1},\dots,k_{t})\in\mathbb{Z}^{t}$, we take a vector $\bm{x}_{\bm{k}}\in\mathbb{R}^{n}$ such that $(\bm{m}_{j},\bm{x}_{\bm{k}})=k_{j}$ $(j=1,\dots,t)$. For $\bm{x}\in\mathbb{R}^{n}$, the condition $(\bm{m}_{j},\sigma^{-1}(\bm{x}))\linebreak=k_{j}$ $(j=1,\dots,t)$ is equivalent to $\sigma^{-1}(\bm{x})-\bm{x}_{\bm{k}}\in\langle\bm{m}_{1},\dots,\bm{m}_{t}\rangle^{\perp}_{\mathbb{R}}$. Therefore we have

[TABLE]

Suppose that $vol({\mathfrak{D}}(f,\sigma))>0$; if the property $vol({\mathfrak{D}}(f,\sigma)\cap{\mathfrak{D}}(f,\mu))>0$ holds, then (47) implies $\langle\sigma(\bm{m}_{1}),\dots,\sigma(\bm{m}_{t})\rangle^{\perp}_{\mathbb{R}}=\langle\mu(\bm{m}_{1}),\dots,\mu(\bm{m}_{t})\rangle^{\perp}_{\mathbb{R}}$, i.e.

[TABLE]

Since the matrix whose $j$th row is $\bm{m}_{j}$ is integral with every elementary divisor being $1$, the above is equivalent to

[TABLE]

hence $\mu^{-1}\sigma\in{\bm{G}}$ holds and the above identity implies $\bm{x}\in\mathfrak{D}(f,\sigma)\iff\bm{x}\in\mathfrak{D}(f,\mu)$ by (46), hence $\mathfrak{D}(f,\sigma)=\mathfrak{D}(f,\mu)$.

∎

Corollary 4.5**.**

We have

[TABLE]

Proof.

Put

[TABLE]

Then we have

[TABLE]

∎

In Proposition 5.2, we show that (48) is equal to $\sqrt{\det((\bm{m}_{i},\bm{m}_{j}))}$ under some conditions.

Lemma 4.2**.**

Suppose that $f(x)$ has no rational root and $\alpha_{i}-\alpha_{j}\notin\mathbb{Q}$ for any distinct $i,j$ and $\#Spl(f,\sigma,\{k_{j}\})=\infty$. For a prime $p\in Spl(f,\sigma,\{k_{j}\})$ with finite exceptions, we have

[TABLE]

and

[TABLE]

where $tr$ denotes the trace from $\mathbb{Q}(f)$ to $\mathbb{Q}$, and $d:=[\mathbb{Q}(f):\mathbb{Q}]$.

Proof.

Since the set

[TABLE]

is an infinite set, the original equation $\sum_{i}m_{j,i}\alpha_{i}=m_{j}$ imply $\sum_{i}m_{j,i}tr(\alpha_{i})=dm_{j}$, hence for every $p\in Spl(f,\sigma,\{k_{j}\})$, we have

[TABLE]

which is the first statement. Next, let us show the second, i.e.

[TABLE]

To prove it, it is enough to show that for $X:=\sum_{i}|tr(\alpha_{i})|/d$, all of conditions $r_{1}>X,p-r_{n}>X,r_{i+1}-r_{i}>X$ $(i=1,\dots,n-1)$ hold for a sufficiently large prime $p$ in $Spl(f,\sigma,\{k_{j}\})$. If the inequality $(0\leq)\,r_{1}\leq X$ occurs for infinitely many primes, then there is an integer $r\in[0,X]$ such that $f(r)\equiv 0\bmod p$ for infinitely many $p$, which means $f(r)=0$. If the inequality $(0<)\,p-r_{n}\leq X$ occurs for infinitely many primes, then there is an integer $r$ such that $r=p-r_{n}\leq X$, hence $f(-r)\equiv 0\bmod p$ for infinitely many $p$, which means $f(-r)=0$. If $r_{i+1}-r_{i}\leq X$ occurs for infinitely many primes, then the equality $r_{i+1}-r_{i}=r(\leq X)$ occurs for some integer $r$ for infinitely many primes $p$, that is polynomials $f(x),f(x+r)\bmod p$ have a common root $r_{i}$ for infinitely many primes, which implies that $f(x),f(x+r)$ are not relatively prime, and they are divisible by some irreducible polynomial $h(x)$, hence $f(x)$ is divisible by $h(x),h(x-r)$, so for a root $\alpha$ of $h(x)$, $\alpha,\alpha+r$ are roots of $f(x)$. This contradicts the assumption. ∎

Proposition 4.5**.**

Suppose that $f(x)$ has no rational root and $\alpha_{i}-\alpha_{j}\notin\mathbb{Q}$ for any distinct $i,j$. Then we see that the condition $\#Spl(f,\sigma,\{k_{j}\})=\infty$ implies $vol(\mathfrak{D}(f,\sigma,\{k_{j}\}))>0$.

Proof.

We note that $\mathfrak{D}(f,\sigma,\{k_{j}\})$ is the intersection of $(n-t)$-dimensional set $\{\bm{x}\in\mathbb{R}^{n}\mid\sum_{i}m_{j,i}x_{\sigma(i)}=k_{j}\,({}^{\forall}j)\}$ and $\{\bm{x}\mid 0\leq x_{1}\leq\dots\leq x_{n}\leq 1\}$. And its $(n-t)$-dimensional open subset $\{\bm{x}\in\mathbb{R}^{n}\mid\sum_{i}m_{j,i}x_{\sigma(i)}=k_{j}\,({}^{\forall}j)\}\cap\{\bm{x}\mid 0<x_{1}<\dots<x_{n}<1\}$ is not empty, since it contains

[TABLE]

by Lemma 4.2. Therefore we have $vol(\mathfrak{D}(f,\sigma,\{k_{j}\}))>0$. ∎

Lastly, let us give the following

Proposition 4.6**.**

We see that

[TABLE]

and, for an integer $k$ $(1\leq k\leq n-1)$,

[TABLE]

Proof.

We note that $vol$ is the $(n-1)$-dimensional volume here. Let $\theta$ be the angle between two hyperplanes defined by $x_{n}=0$ and $\sum x_{i}=0$, that is the angle between vectors $(0,\dots,0,1)$ and $(1,\dots,1)$, hence $\cos\theta=1/\sqrt{n}$. Since a permutation acts on $\mathbb{R}^{n}$ and on the set $\{\bm{x}\in\mathbb{R}^{n}\mid\sum x_{i}=k\}$ as an orthogonal transformation and the dimension of the set defined by $\sum x_{i}\in\mathbb{Z}$ and $x_{i}=x_{i+1}$ is $n-2$, $vol(\hat{\mathfrak{D}}_{n})$ is equal to the volume of the set

[TABLE]

hence we see that

[TABLE]

and the projection $(x_{1},\dots,x_{n})\mapsto(x_{1},\dots,x_{n-1})$ from $\{(x_{1},\dots,x_{n})\in[0,1)^{n}\mid\sum x_{i}\in\mathbb{Z}\}$ to $[0,1)^{n-1}$ is bijective by $x_{1}+\dots+x_{n-1}+x_{n}=\lceil x_{1}+\dots+x_{n-1}\rceil$, we have $n!vol(\hat{\mathfrak{D}}_{n})\cos\theta=vol([0,1)^{n-1})=1.$ Next, we have

[TABLE]

∎

**Remark ** The set $\hat{\mathfrak{D}}_{n}$ is in a union of parallel transformations of the subspace $S:=\{\bm{x}\in\mathbb{R}^{n}\mid\sum_{i=1}^{n}x_{i}=0\}$. Write $\bm{e}=(1,\dots,1),\bm{f}=(0,\dots,0,\sqrt{n})$ and let $\tau$ be the symmetry defined by $\bm{x}\mapsto\bm{x}-\frac{2(\bm{e}-\bm{f},\bm{x})}{(\bm{e}-\bm{f},\bm{e}-\bm{f})}(\bm{e}-\bm{f})$; then by $S=\bm{e}^{\perp}=\tau(\bm{f})^{\perp}$, we see that $\tau(\bm{e}_{i})=\bm{e}_{i}-\frac{1}{n-\sqrt{n}}(\bm{e}-\bm{f})$ $(i=1,\dots,n-1)$ are orthnormal basis of $S$, where $\bm{e}_{1}=(1,0,\dots,0),\bm{e}_{2}=(0,1,0,\dots,0)$ and so on. Another basis of $S$ is $\bm{u}_{i}:=\frac{1}{\sqrt{(n-i)(n-i+1)}}(0,\dots,0,n-i,-1,\dots,-1)$, where the number $n-i$ is on the $i$th entry.

Let us give a geometric remark here. For linearly independent vectors $\bm{u}_{1},\dots,\linebreak[3]\bm{u}_{T}\in\mathbb{R}^{n}$ $(1\leq T<n)$, write

[TABLE]

and we add vectors $\bm{u}_{T+1},\dots,\bm{u}_{n}$ so that the matrix U=\left(\begin{array}[]{cc}U_{1}^{(T)}&U_{2}\\ U_{3}&U_{4}\end{array}\right) with $i$th row being $\bm{u}_{i}$ is regular. Moreover, let vectors $\bm{f}_{i}\in\mathbb{R}^{n}$ $(i=1,\dots,T)$ be orthonormal basis of $S$ and define the matrix $C=(c_{ij})$ by $\bm{u}_{i}=\sum_{j=1}^{T}c_{ij}\bm{f}_{j}$.

Defining the projection $pr$ from $\mathbb{R}^{n}$ to $V:=\{(x_{1},\dots,x_{T},0,\dots,0)\mid x_{i}\in\mathbb{R}\}$ by

[TABLE]

let us show that there is a constant $c_{1}$ such that ${vol(pr(Y))}=c_{1}{vol(Y)}$ for any measurable set $Y$ on $S$ under the condition $\det U_{1}\neq 0$.

First, let us find the condition to $pr(S)=V$. Suppose that $\bm{y}=(y_{1},\dots,y_{n})\in\mathbb{R}^{n}$ and $\bm{x}=(x_{1},\dots,x_{T})\in\mathbb{R}^{T}$ satisfy the relation $\bm{y}=\sum_{i=1}^{T}x_{i}\bm{u}_{i}$, which is equivalent to

[TABLE]

by $\bm{y}=(y_{1},\dots,y_{n})=(x_{1},\dots,x_{T},0,\dots,0)U$. Hence the conditions $pr(S)=V$ and $\det U_{1}\neq 0$ are equivalent, and then the mapping $pr$ is the isomorphism from $S$ to $V$ and we see

[TABLE]

Write $Y:=\{\sum_{i=1}^{T}x_{i}\bm{u}_{i}\mid(x_{1},\dots,x_{T})\in X\}$ for a measurable set $X$ in $\mathbb{R}^{T}$. Then it is easy to see

[TABLE]

hence $vol(Y)=vol(\{(x_{1},\dots,x_{T})C\mid(x_{1},\dots,x_{T})\in X\}$ by our definition and it is equal to $|\det C|vol(X)$. Therefore we see

[TABLE]

Thus the constant $c_{1}=|\det U_{1}||\det C|^{-1}$ which is independent of $Y$ satisfies $vol(pr(Y))=c_{1}vol(Y)$.

In case of $Y=\{\sum_{i=1}^{T}x_{i}\bm{u}_{i}\mid 0\leq x_{i}\leq 1\,(i=1,\dots,T)\}$, we see $vol(Y)=\sqrt{\det((\bm{u}_{i},\bm{u}_{j}))}=|\det{C}|$ and $vol(pr(Y))=|\det{U_{1}}|$.

In case of $Y=\{\sum_{i=1}^{T}x_{i}\bm{f}_{i}\mid 0\leq x_{i}\leq 1(i=1,\dots,T)\}$, we see $vol(Y)=1$ and $vol(pr(Y))=$ the absolute value of the determinant of the left $(T,T)$-submatrix of \left(\begin{array}[]{c}\bm{f}_{1}\\ \vdots\\ \bm{f}_{T}\end{array}\right).

As an example, let us see the subspaces $S=\{\bm{x}\in\mathbb{R}^{n}\mid\sum_{i=1}^{n}x_{i}=0\}$, and $V=\{\bm{x}\in\mathbb{R}^{n}\mid x_{n}=0\}$. For $S$, we take the orthonormal basis $\bm{u}_{i}=\bm{f}_{i}:=\frac{1}{\sqrt{(n-i)(n-i+1)}}(0,\dots,0,n-i,-1,\dots,-1)$ $(i=1,\dots,n-1)$ given in the remark after Proposition 4.6. Then, writing the $(i,j)$-entry of the matrix $U_{1}$ by $u_{i,j}$, we see that $u_{i,j}=0$ if $i>j$ and $u_{i,i}=\sqrt{\frac{n-i}{n-i+1}}$, hence $c_{1}=\det U_{1}=\frac{1}{\sqrt{n}}$.

5 One-dimensional distribution

In this section, we discuss the one-dimensional equi-distribution of $r_{i}/p$ $(1\leq i\leq n,p\in Spl(f))$ for local roots $r_{i}$, that is whether, for $0\leq a<1$

[TABLE]

is true or not. We note that if $f(x)$ has no rational integral root, then the number of primes $p\in Spl(f)$ satisfying $r_{1}<X$ or $r_{n}>p-X$ is finite for any number $X$, and if $f(x)$ has a rational integral root $r_{0}$, then $r=r_{0}$ or $r=p+r_{0}$ is a local root according to $r_{0}\geq 0$ or $r_{0}<0$ if $p\in Spl(f)$ is large, hence $r/p$ tends to [math] or $1$ and the one-dimensional equi-distribution is false. So, we exclude such polynomials. In 5.1, we treat the case that a polynomial $f(x)$ has no non-trivial linear relation among roots, and in 5.2 the case of a decomposable and irreducible polynomial of degree 4. In 5.3, for a more general polynomial, e.g. an irreducible polynomial of degree larger than $1$, we show that Conjectures 1, 2 imply the one-dimensional equi-distribution of roots $r_{i}/p$ not based on hard calculation.

Supposing $f(0)\neq 0$, we see easily that for a rational number $a$, there are only finitely many primes $p\in Spl(f)$ such that $a-r/p=O(\frac{1}{ph(p)})$ where $r$ is a local root and $h(x)$ is any function satisfying $h(x)\to\infty$ $(x\to\infty)$.

5.1 Case without non-trivial linear relation among roots

The aim in this subsection is to prove the following two theorems.

Theorem 5.1**.**

Let $f(x)$ be a polynomial of degree $n\,(>1)$, which has no non-trivial linear relation among roots. Then the conjecture (11) implies for $1\leq i\leq n$, $0\leq a<1$

[TABLE]

where the binomial coefficient $A\choose B$ is supposed to vanish unless $0\leq B\leq A$, and $M(x):=\max(x,0).$

When $i=1$, a simpler formula is given in Proposition 5.1.

Example Denoting $\lim_{X\to\infty}\frac{\#\{p\in Spl_{X}(f)\mid r_{i}/p<a\}}{\#Spl_{X}(f)}$ by $D(a,i;n)$, we see

[TABLE]

Using Theorem 5.1, we show the one-dimensional equi-distribution of $r_{i}/p$.

Theorem 5.2**.**

Let $f(x)$ be a polynomial of degree $n\,(>1)$ which has no non-trivial linear relation among roots; then (11) implies

[TABLE]

for a real number $a\in[0,1)$.

In case of $n=2$, the equation (49) is proved by [DFI] and [T].

The rest of this subsection is devoted to their proof and let us start from the following easy lemma.

Lemma 5.1**.**

For $t,a\in\mathbb{R}$, we have

[TABLE]

Proof.

The left-hand side is equal to

[TABLE]

∎

For the sake of convenience, we recall the proof of the following [Fe].

Lemma 5.2**.**

For a natural number $k$, the volume $U_{k}(x)$ $(x\in\mathbb{R})$ of a subset of the unit cube $[0,1)^{k}$ defined by $\{(x_{1},\dots,x_{k})\in[0,1)^{k}\mid x_{1}+\dots+x_{k}\leq x\}$ is given by

[TABLE]

Proof.

In case of $k=1$, $U_{1}(x)=M(x)-M(x-1)$ is easy, hence the lemma is true. Suppose that the lemma is true for $k\geq 1$; then we see that

[TABLE]

which completes the induction. ∎

Corollary 5.1**.**

Let $n\,(\geq 1)$ be an integer. Then we have, for a polynomial $P(x)=c_{n}x^{n}+\dots+c_{0}$, we have

[TABLE]

Proof.

The case of $n=1$ is obvious. The equation (51) means

[TABLE]

The equation (50) implies that $(U_{n}(x)=)\frac{1}{n!}\sum_{k=0}^{n}(-1)^{k}{n\choose k}M(x-k)^{n}=1$ if $x>n$, hence we have the identity $\sum_{k=0}^{n}(-1)^{k}{n\choose k}(x-k)^{n}=n!$. We find that the coefficient of $x^{n-j}$ of the identity is $\sum_{k=0}^{n}(-1)^{k}{n\choose k}{n\choose j}(-k)^{j}$. If, hence $j<n$, then we have $\sum_{k=0}^{n}(-1)^{k}{n\choose k}k^{j}=0$, i.e. (51) for a polynomial $P$ with $\deg P<n$. If $j=n$, then $\sum_{k=0}^{n}(-1)^{k}{n\choose k}(-k)^{n}=n!$ which implies (51) for $P(x)=x^{n}$. ∎

Corollary 5.2**.**

For natural numbers $k,n$ with $1\leq k\leq n$ we have

[TABLE]

where $A(n,k)$ denotes the Eulerian number in the introduction.

Proof.

The first equality follows from

[TABLE]

Putting $B(n,k):=n!(U_{n}(k)-U_{n}(k-1))$, we have only to confirm

[TABLE]

The equation $B(1,1)=1$ is clear, and we see that

[TABLE]

∎

Let $f(x)$ be a polynomial of degree $n\,(>1)$ without non-trivial linear relation among roots. Given number $a\in[0,1]$, putting

[TABLE]

we see that Conjecture 2 implies

[TABLE]

hence to prove Theorem 5.2, we have only to show that the above value is equal to $a$. Hereafter, a real number $a$ with $0\leq a<1$ is fixed.

Lemma 5.3**.**

For an integer $k$ with $1\leq k\leq n$, let

[TABLE]

for the angle $\theta$ of two hyper-planes defined by $x_{j}=0$ and by $x_{1}+\dots+x_{n}=0$ in $\mathbb{R}^{n}$. Then we have

[TABLE]

Proof.

It is easy to see

[TABLE]

using $vol(\hat{\mathfrak{D}}_{n})=\frac{1}{n!\cos\theta}$. ∎

Lemma 5.4**.**

For $k=n$, we have

[TABLE]

Proof.

It is easy to see that

[TABLE]

The volume of the set $\{\bm{x}\in\mathbb{R}^{n-1}\mid 0\leq x_{1},\dots,x_{n-1}\leq a,\sum_{i=1}^{n-1}x_{i}\leq K\}$ is equal to

[TABLE]

Therefore we have

[TABLE]

∎

Lemma 5.5**.**

In case of $1\leq k\leq n-1$, we have

[TABLE]

where

[TABLE]

Proof.

We see that

[TABLE]

∎

Lemma 5.6**.**

For integers $r,k$ with $1\leq k\leq r$, we have

[TABLE]

Proof.

This follows from the equation

[TABLE]

where the domain $D$ is given by the conditions

[TABLE]

∎

Lemma 5.7**.**

For integers $j,k$ with $j\geq 0,k\geq 1$, $U_{k,k+j}(t)$ is equal to

[TABLE]

Proof.

Suppose that $j=0$; then $U_{k,k}(t)$ equals

[TABLE]

Second, suppose that the equation (60) is true; then we see that $U_{k,k+j+1}(t)$ equals

[TABLE]

which completes the induction. ∎

Lemma 5.8**.**

For $1\leq k\leq n-1$, we have

[TABLE]

Proof.

For $1\leq k,m\leq n-1$, Lemma 5.7 implies

[TABLE]

noting the supposition that the binomial coefficient $A\choose B$ vanishes unless $0\leq B\leq A$. Continuing the calculation, it is

[TABLE]

hence we have

[TABLE]

where the restrictions on $r,s$ follow from conditions $1\leq k,m\leq n-1,0\leq n+r-s-m\leq k,0\leq m-r\leq n-k$. Lemma 5.5 completes the proof. ∎

Lemma 5.9**.**

Let $m,n$ be integers satisfying $0\leq m\leq n-1$. Then we have

[TABLE]

This is well-known and we omit the proof.

Proposition 5.1**.**

For an integer $i$ with $1\leq i\leq n$ and a real number $a\in[0,1)$, we have

[TABLE]

In particular, we have for $i=1$

[TABLE]

where

[TABLE]

Proof.

By Lemmas 5.3, 5.4, 5.8, we have

[TABLE]

since the binomial coefficient ${0\choose m-l}$ vanishes unless $m=l$. The partial sum $\sum_{m=1}^{n-1}{k\choose n-h-m+l}{n-k\choose m-l}$ is equal to

[TABLE]

where we use for the second equality, the formula $\sum_{r=0}^{p}{n\choose r}{m\choose p-r}={n+m\choose p}$ if $p\leq n+m$.

Next, let us assume $i=1$ to show (62). Putting

[TABLE]

we have only to prove $T(l,h)=(n-1)!\,C_{1}(l,h)$. It is obviously true if $h\geq l+1$, since the partial sum on $q$ is empty. In case of $h=0$, we see that $T(l,0)$ is equal to

[TABLE]

by Lemma 5.9. Lastly assume that $1\leq h\leq l$. The sum $T(l,h)+(-1)^{h+n}{n\choose h}$ is equal to

[TABLE]

which implies $T(l,h)=0$. ∎

The proposition gives Theorem 5.1 by (63), and we see that (54) is the sum of $C(l,h)M(l-ha)^{n-1}$ over integers $l,h$ satisfying

[TABLE]

where

[TABLE]

using Corollary 5.1 with $P(x)=x$ and $n>1$. To prove (54) being to equal to $a$, we will show

[TABLE]

Under the equations (66), (54) being $a$, i.e. Theorem 5.1 is proved as follows: The expression (54) is equal to

[TABLE]

by Corollary 5.1.

Let us show (66). Suppose $h=0$; we see that

[TABLE]

which is (66).

Second we see that

[TABLE]

Unless $q-1\leq k$ and $n-q\leq n-k$, binomial coefficients vanish, hence we may assume that $q=k$ or $q=k+1$, and we see

[TABLE]

Finally, assume that $h\geq 2$; hence $1\leq l\leq n-1,2\leq h\leq n$ are supposed. By (65), we have

[TABLE]

since

[TABLE]

using Corollary 5.1 to $P(x)=q-x$ with $h\geq 2$. Thus we have completed the proof of Theorem 5.2.

5.2 Case of degree $4$ with non-trivial linear relation among roots

Even if an irreducible polynomial has non-trivial linear relations among roots, the one-dimensional equi-distribution of $r_{i}/p$ for a local root is likely. The following is an example.

Theorem 5.3**.**

For an irreducible polynomial $f(x)$ of degree $4$ with non-trivial linear relation among roots, Conjecture 2 implies the equi-distribution of $r_{i}/p$ $(1\leq i\leq 4,p\in Spl(f))$.

Proof.

Let an irreducible polynomial $f(x)$ of degree $4$ with non-trivial linear relation among roots be $f(x)=(x^{2}+ax)^{2}+b(x^{2}+ax)+c$ by Proposition 3.2; then by defining numbers $\beta_{i},\alpha_{i,j}$ by

[TABLE]

as Proposition 3.3, a basis of linear relations among roots is $\alpha_{i,1}+\alpha_{i,2}=-a$ $(i=1,2)$. Letting $\alpha_{1}:=\alpha_{1,1},\alpha_{2}:=\alpha_{2,1},\alpha_{3}:=\alpha_{2,2},\alpha_{4}:=\alpha_{1,2}$, we have $\alpha_{1}+\alpha_{4}=\alpha_{2}+\alpha_{3}=-a$, hence $t=2$ and $\hat{\bm{m}}_{1}=(1,0,0,1,-a),\hat{\bm{m}}_{2}=(0,1,1,0,-a)$. For local solutions $r_{i}$ for a large prime $p\in Spl(f)$, induced linear relations in (7) are $r_{1}+r_{4}=r_{2}+r_{3}=-a+p$, i.e. $k_{1}=k_{2}=1$, and $Spl_{X}(f,\sigma)\neq\emptyset$ if and only if $\{\sigma(1),\sigma(4)\}=\{1,4\}$ or $\{2,3\}$. Thus, we see $\#\hat{G}=8,[\mathbb{Q}(f):\mathbb{Q}]=4,8$. For such $\sigma,k_{j}$, we have $Spl_{X}(f)=Spl_{X}(f,\sigma)$ with possibly finitely many exceptional primes. Only for such permutations, $vol(\mathfrak{D}(f,\sigma))>0$ and $\mathfrak{D}(f,\sigma)=\mathfrak{D}(f,id)$ are easy, hence we have, by (11)

[TABLE]

Hence we have only to show

[TABLE]

by putting $D_{i}:=\{(x_{1},\dots,x_{4})\mid x_{i}\leq A\}\cap\,\mathfrak{D}(f,id)$. By $\mathfrak{D}(f,id)=\{(x_{1},\dots,x_{4})\mid 0\leq x_{1}\leq\dots\leq x_{4}<1,x_{1}+x_{4}=1,x_{2}+x_{3}=1\}$ we have

[TABLE]

and projecting them on the $(x_{1},x_{2})$-plane, we see

[TABLE]

Thus we see that (68) is true. ∎

5.3 More general case

In this subsection, assume Conjectures 1, 2 and that $f(x)$ has no rational root and $\alpha_{i}-\alpha_{j}\not\in\mathbb{Q}$ if $i\neq j$, and the conditions $Pr(f,\sigma)>0,vol(\mathfrak{D}(f,\sigma))>0$ are equivalent. Using Corollary 4.2, we have a similar expression to (54) with the definition (53) of $D_{i,a}$ for a more general polynomial $f(x)$:

[TABLE]

We note that the dimension of the set

[TABLE]

is less than $n-t$, since the condition $x_{i}=0$ is independent of conditions $\sum_{k=1}^{n}m_{j,k}x_{\sigma(k)}\in\mathbb{Z}\,({}^{\forall}j)$ by the assumption that $f(x)$ has no rational root. Therefore the above is equal to [math] at $a=0$, and to $1$ at $a=1$ by

[TABLE]

Hence, once we have shown that it is a linear form in $a$, it is equal to $a$ on $[0,1)$, that is the distribution of $r_{i}/p$ is uniform under the above conditions. Till here, we do not use the assumption that $vol(\mathfrak{D}(f,\sigma))>0$ implies $Pr(f,\sigma)>0$ and the difference between the sum restricted on $S_{n}^{\prime}$ above and the full sum

[TABLE]

is

[TABLE]

From now on, the aim is to show that (69) is a linear form in $a$, which means that the distribution of $r_{i}/p$ is uniform. If, moreover (67) is also a linear form in $a$, then the distribution of $r_{i}/p$ is uniform without the assumption that ${vol}(D_{i,a}\cap{\mathfrak{D}}(f,\sigma))>0$ implies $Pr(f,\sigma)>0$. (cf. 6.4.2).

Now we assume that a polynomial $f(x)$ has no rational roots, and put

[TABLE]

Recall that vectors $\bm{m}_{1},\dots,\bm{m}_{t}$ are a basis of $LR_{0}\cap\mathbb{Z}^{n}$, hence $\mathfrak{D}_{f}$ is $G$-stable.

Lemma 5.10**.**

Suppose that $\alpha_{i}-\alpha_{j}\not\in\mathbb{Q}$ for any distinct $i,j$, moreover. We have

[TABLE]

in particular,

[TABLE]

Proof.

The equations

[TABLE]

imply

[TABLE]

Here, we note that $\mathfrak{D}_{f}$ is defined by $t$ linearly independent vectors $\bm{m}_{j}$ and the volume $vol$ is $(n-t)$-dimensional one, hence it is not necessary to care an additional equation $x_{i}=x_{j}$ for $i\neq j$ by the assumption that the vector space $LR_{0}$ spanned by $\bm{m}_{j}$ does not contain vectors $(0,\dots,0,1,0,\dots,0,-1,0,\dots,0)$. Let us define a mapping $\phi$ from

[TABLE]

to the union of

[TABLE]

by $\phi((\bm{x},i))=(\bm{x},\sigma,k)$, where $\sigma,k$ are defined by

[TABLE]

hence $\phi(X(i))=\cup_{\sigma,k\text{:}\sigma(k)=i}Y(\sigma,k)$. The mapping is clearly bijective, and if $(\bm{y},\sigma,k)\in Y(\sigma,k)$ and $(\bm{y},\sigma^{\prime},k^{\prime})\in Y(\sigma^{\prime},k^{\prime})$ with $\sigma(k)=\sigma^{\prime}(k^{\prime})$ occur, then $\sigma=\sigma^{\prime},k=k^{\prime}$ hold, hence, defining $pr$ by $pr(\bm{y},\sigma,k)=\bm{y}$, we have $pr(Y(\sigma,k))\cap pr(Y(\sigma^{\prime},k^{\prime}))=\emptyset$ if $\sigma(k)=\sigma^{\prime}(k^{\prime})$ and either $\sigma\neq\sigma^{\prime}$ or $k\neq k^{\prime}$, and so

[TABLE]

∎

We note that

[TABLE]

and putting

[TABLE]

it is easy to see that the mapping $\phi$ from the disjoint union $\sqcup_{\sigma}{\mathfrak{D}}(f,\sigma)^{\circ}$ to $(0,1)^{n}\cap\mathfrak{D}_{f}^{\circ}$ defined by $\phi(\bm{x})=(x_{\sigma(1)},\dots,x_{\sigma(n)})$ for $\bm{x}\in{\mathfrak{D}}(f,\sigma)^{\circ}$ is bijective. (cf. Proposition 4.4)

To show the equi-distribution, we have only to see that each factor ${vol}(\{\bm{x}\in[0,1)^{n}\mid x_{i}\leq a\}\cap{\mathfrak{D}}_{f})$ is a linear form in $a$.

Lemma 5.11**.**

Suppose that $\beta_{1},\dots,\beta_{r}$ are linearly independent over $\mathbb{Q}$ and

[TABLE]

Then we have $r=n-t$,

[TABLE]

and

[TABLE]

Proof.

By

[TABLE]

we see

[TABLE]

hence $t=\dim{LR_{0}}=n-r$. On the other hand, the definition of $T$ implies

[TABLE]

hence row vectors of the matrix $({}^{t}T,-1_{n-r})$ span $LR_{0}$ by $\text{rank}({}^{t}T,-1_{n-r})=\dim LR_{0}$ and so there is a matrix $S\in GL_{n-r}(\mathbb{Q})$ such that

[TABLE]

∎

Lemma 5.12**.**

Supposing the assumption on $\beta_{i}$ in Lemma 5.11, we have

[TABLE]

For $(x_{1},\dots,x_{r})\in[0,1)^{r}$, there are exactly $|\det(S)|$ vectors $(x_{r+1},\dots,x_{n})\in[0,1)^{n-r}$ such that $(x_{1},\dots,x_{n})\in\mathfrak{D}_{f}$.

Proof.

We see that

[TABLE]

Since $S\in GL_{n-r}(\mathbb{Q})$ is an integral regular matrix, the inclusion $\mathbb{Z}^{n-r}\,{}^{t}\!S^{-1}\supset\mathbb{Z}^{n-r}$ is clear. Hence, for $(x_{1},\dots,x_{r})\in\mathbb{R}^{r}$, there is a vector $(x_{r+1},\dots,x_{n}):=(x_{1},\dots,x_{r})T+\bm{k}{}^{t}S^{-1}\in[0,1)^{n-r}$ for some integral vector $\bm{k}$. The number of such integral vectors is $|\det(S)|$ since $S$ is integral. ∎

Therefore we see that, assuming that $\beta_{1},\dots,\beta_{r}$ is linearly independent over $\mathbb{Q}$ and $1\leq i\leq r$ as above

[TABLE]

which is included in the union of sets parallel to the subspace

[TABLE]

and the projection to $\{\bm{x}\in[0,1)^{r}\mid x_{i}\leq a\}$ is $|\det(S)|$-fold on by the lemma above. Hence the volume of $\{\bm{x}\in[0,1)^{n}\mid x_{i}\leq a\}\cap\mathfrak{D}_{f}$ as an $r\,(=n-t)$-dimensional set is proportional to $a$. In general, for given $i$, we have only to take a subset $j_{1},\dots,j_{r}$ such that $\beta_{j_{1}},\dots,\beta_{j_{r}}$ are linearly independent over $\mathbb{Q}$ and $j_{k}=i$ for some $k$. Thus, we have proved

Theorem 5.4**.**

If a polynomial $f(x)$ has no rational root, $\alpha_{i}-\alpha_{j}\not\in\mathbb{Q}$ holds for any distinct $i,j$, and the conditions $Pr(f,\sigma)>0$ and ${vol}({\mathfrak{D}}(f,\sigma))>0$ are equivalent, then Conjectures 1, 2 imply the equi-distribution of $r_{i}/p$ for local roots $r_{i}$ of the polynomial.

So, the equi-distribution of $r_{i}/p$ for local roots $r_{i}$ is likely for an irreducible polynomial $f(x)$ of $\deg f>1$. Although the polynomial $f(x)=(x^{2}-2)((x-1)^{2}-2)$ does not satisfy the assumption of the theorem, the equi-distribution of local roots $r_{i}/p$ is true (cf. 6.4.2).

Put, for an algebraic number field $F$ containing $\mathbb{Q}(f)$

[TABLE]

If a subsequence of $r_{i}/p$ for local roots $r_{i}$ of $f$ on $p\in Spl(f,F)$ distributes uniformly for every irreducible polynomial $f$ of $\deg f>1$ and every number field $F$, then the one-dimensional distribution of $r_{i}/p$ for any reducible polynomial without rational root is true.

Let us give more remarks.

Proposition 5.2**.**

Suppose that $f(x)$ is not a product of linear forms and $\alpha_{i}-\alpha_{j}\not\in\mathbb{Q}$ for any distinct $i,j$; then we have

[TABLE]

Proof.

The first two equalities are nothing but (48) and (72). The matrix whose rows are $\bm{m}_{1},\dots,\bm{m}_{t}$ is primitive, hence there are integral vectors $\bm{m}_{t+1},\dots,\bm{m}_{n}$ such that the matrix $M$ with the $i$th row being $\bm{m}_{i}$ is unimodular, that is $M$ is an integral matrix with determinant $1$. Denote the $i$-th row of the matrix ${}^{t}\!M^{-1}$ by $\bm{m}^{\prime}_{i}$, hence $(\bm{m}^{\prime}_{i},\bm{m}_{j})=\delta_{i,j}$. We see $\bm{x}+\bm{m}\in\mathfrak{D}_{f}$ for $\bm{x}\in\mathfrak{D}_{f}$ and $\bm{m}\in\mathbb{Z}^{n}$, hence $\mathbb{Z}^{n}$ acts on $\mathfrak{D}_{f}$ as a volume-preserving discontinuous group. The set $[0,1)^{n}\cap\mathfrak{D}_{f}$ is a fundamental domain ( $=$ a realization of $\mathfrak{D}_{f}/\mathbb{Z}^{n}$). Another is $\{\sum_{i=t+1}^{n}x_{i}\bm{m}^{\prime}_{i}\mid{}^{\forall}x_{i}\in[0,1)\}$, since for a vector $\bm{x}=\sum x_{i}\bm{m}^{\prime}_{i}$ in

[TABLE]

the condition $\bm{x}\in\mathbb{Z}^{n}$ holds if and only if $x_{t+1},\dots,x_{n}\in\mathbb{Z}$ holds. Hence we have $vol([0,1)^{n}\cap\mathfrak{D}_{f})=vol(\{\sum_{i=t+1}^{n}y_{i}\bm{m}^{\prime}_{i}\mid 0\leq{}^{\forall}y_{i}<1\})=\sqrt{\det((\bm{m}^{\prime}_{,}\bm{m}^{\prime}_{j}))_{t<i,j\leq n}}$. Decompose the matrix $M{}^{t}M$ as

[TABLE]

which implies $\det A\cdot\det(C-{}^{t}BA^{-1}B)=1$. Here we note that $M{}^{t}\!M$ is positive definite, and $A=((\bm{m}_{i},\bm{m}_{j}))_{1\leq i,j\leq t}$. Similarly, the right lower $(n-t)$-square matrix of ${}^{t}\!M^{-1}{}^{t}({}^{t}\!M^{-1})=(M{}^{t}\!M)^{-1}$ is $((\bm{m}^{\prime}_{i},\bm{m}^{\prime}_{j}))_{t+1\leq i,j\leq n}$. By the above decomposition, we see that $(M{}^{t}\!M)^{-1}$ is

[TABLE]

hence we see that $\det((\bm{m}^{\prime}_{i},\bm{m}^{\prime}_{j}))_{t+1\leq i,j\leq n}=\det(C-{}^{t}BA^{-1}B)^{-1}=\det A=\det((\bm{m}_{i},\bm{m}_{j}))_{1\leq i,j\leq t}$, which completes the proof. ∎

The equation above is a little bit curious, because $vol$ is, by definition the $(n-t)$-dimensional volume and $\sqrt{\det((\bm{m}_{i},\bm{m}_{j}))}$ is the volume of the $t$-dimensional paralleltope spanned by $\bm{m}_{1},\dots,\bm{m}_{t}$.

Proposition 5.3**.**

Suppose that $f(x)$ is not a product of linear forms, $\alpha_{i}-\alpha_{j}\not\in\mathbb{Q}$ for any distinct $i,j$ and $vol(\mathfrak{D}(f,\sigma))=cPr(f,\sigma)$ for every $\sigma\in S_{n}$ ; then we have

[TABLE]

Proof.

We see easily

[TABLE]

hence we have

[TABLE]

∎

**Example ** The case that the polynomial $f(x)$ has no non-trivial linear relation among roots : It is easy to see $Pr(f,\sigma)=1$ and $vol(\mathfrak{D}(f,\sigma))=vol(\hat{\mathfrak{D}}_{n})=\frac{\sqrt{n}}{n!}$ by Proposition 4.6, hence $c=\frac{\sqrt{n}}{n!}$. On the other hand, the base of $LR_{0}\cap\mathbb{Z}^{n}$ is $(1,\dots,1)$ and $\hat{G}=S_{n}$ is clear, hence the identity in Proposition 5.3 is checked.

The case that the irreducible polynomial $f(x)$ is of degree $4$ and has non-trivial linear relations among roots : Suppose that $f(x)$, roots $\alpha_{i}$, the basis $\bm{m}_{i}$ $(i=1,2)$ are those in the proof of Theorem 5.3; then, only for permutations $\sigma$ satisfying $\{\sigma(1),\sigma(4)\}=\{1,4\}$ or $\{2,3\}$, $Pr(f,\sigma)>0,vol(\mathfrak{D}(f,\sigma))>0$ hold, and then $Pr(f,\sigma)=1$ holds and $\mathfrak{D}(f,\sigma)$ is equal to

[TABLE]

hence $vol(\mathfrak{D}(f,\sigma))=1/4$. On the other hand, it is easy to see $\det((\bm{m}_{i},\bm{m}_{j}))=4,\#\hat{G}=8$, hence the identity on $c$ is checked.

6 The Weyl criterion

The equation (13) in Conjecture 3 is equivalent to

[TABLE]

where $F=\chi_{D}$ is the characteristic function of the set $D$ there. Here we remark that for a prime $p\in Spl_{X}(f,\sigma,\{k_{j}\})$, the point $(\frac{r_{1}}{p},\dots,\frac{r_{n}}{p})$ is not on $\mathfrak{D}(f,\sigma,\{k_{j}\})$ unless $m_{1}=\dots=m_{t}=0$ by the condition $\sum_{i}m_{j,i}r_{\sigma(i)}=m_{j}+k_{j}p$, hence the domain of the definition of the function $F$ should be paid attention to. As the original case, we want to approximate the characteristic function by manageable functions, more explicitly trigonometric ones to prove (73).

Let us recall the original Weyl criterion for a uniformly distributed sequence modulo $1$ (see [K-N]): The Weyl criterion states that a sequence $\{x_{n}\}$, $x_{n}\in[0,1)$, $n=1,2,\dots$ is uniformly distributed modulo $1$ if and only if the following equation

[TABLE]

holds for $F(x)=\exp(2\pi ihx)$ $({}^{\forall}h\in\mathbb{Z})$. Usually, the right-hand side is replaced by “[math] for $h\neq 0$”, omitting the trivial case $h=0$. The proof proceeds as follows:

The sequence $\{x_{n}\}$, $x_{n}\in[0,1)$, $n=1,2,\dots$ is uniformly distributed modulo $1$ if and only if the equation (W) holds for the characteristic function of every interval in $[0,1]$, compactifing the interval $[0,1)$ to $[0,1]$ with the identification of endpoints $0,1$. 2. 2.

Approximating a continuous function $F(x)$ $(F(0)=F(1))$ on $[0,1]$ by step functions from above and below by the supremum norm, one proves that the equation (W) is true for step functions if and only if it is true for continuous functions . 3. 3.

Approximating a continuous function $F(x)$ $(F(0)=F(1))$ on $[0,1]$ by trigonometric functions $\exp(2\pi ihx)$ $(h\in\mathbb{Z})$, one shows that (W) is true for continuous functions $F(x)$ if and only if (W) is true for trigonometric functions $\exp(2\pi ihx)$ $({}^{\forall}h\in\mathbb{Z})$.

We want to use the Stone-Weierstrass theorem instead of the approximation by Fourier series to apply the idea to our case.

6.1 The Stone-Weierstrass theorem

The Stone-Weierstrass theorem is the following.

Theorem 6.1**.**

Let $T$ be a compact set and let $C(T)$ be the algebra of $\mathbb{R}$-valued continuous functions on $T$. Suppose that $\mathfrak{A}$ is a subalgebra of $C(T)$ such that (i) $\mathfrak{A}$ contains every constant function, (ii) for distinct points $x,y\in T$, there is a function $g\in\mathfrak{A}$ such that $g(x)\neq g(y)$. Then a continuous function $f$ on $T$ is approximated by functions in $\mathfrak{A}$ by the supremum norm $||f||:=\sup_{x\in T}|f(x)|$.

The proof in [BD] is simple. Just to make sure, we give the following variation.

Corollary 6.1**.**

Let ${T}$ be a compact set and let $E_{\gamma}\,(\gamma\in\Gamma)$ be mutually disjoint subsets of ${T}$, and let $C({T})$ be the algebra of $\mathbb{R}$-valued continuous functions on ${T}$. Suppose that $\mathfrak{A}$ is a subalgebra of $C({T})$ such that (i) $\mathfrak{A}$ contains every constant function, (ii) every function in $\mathfrak{A}$ takes the same value on each $E_{\gamma}$, and (iii) for distinct points $x,y\in{T}$, there is a function $g\in\mathfrak{A}$ such that $g(x)\neq g(y)$ if there is no $\gamma\in\Gamma$ such that $x,y\in E_{\gamma}$. Then a continuous function $f$ on ${T}$ such that $f(x)$ takes the same value on each $E_{\gamma}$ is approximated by functions in $\mathfrak{A}$ by the supremum norm $||f||:=\sup_{x\in{T}}|f(x)|$.

The proof is easy. We have only to apply the Stone-Weierstrass theorem to the quotient space of $T$ by the equivalence shrinking $E_{\gamma}$ to one point for each $\gamma\in\Gamma$.

6.2 The Weyl criterion

Hereafter we suppose that the polynomial $f(x)$ has no rational root and is not of the form $g(x)g(x-c)h(x)$ $(g(x),h(x)\in\mathbb{Z}[x],c\in\mathbb{Z})$, and write

[TABLE]

In view of the Stone-Weierstrass theorem, we want to find a good system of continuous functions to approximate the characteristic function $\chi_{D}$ at (73). So, we want a system of manageable functions $F$ which satisfies (73) in the case of $\#Spl(f,\sigma,\{k_{j}\})=\infty$, which implies $\dim\mathfrak{D}=n-t$ by Proposition 4.5.

Let us take as manageable functions trigonometric ones, and let us evaluate the integral of the right-hand side at (73) for them.

First, let us give another description of $\mathfrak{D}$. Since we see, via

[TABLE]

[TABLE]

with $\bm{w}_{1}:=(1,\dots,1),\bm{w}_{2}:=(0,1,\dots,1),\dots,\bm{w}_{n}:=(0,\dots,0,1)$, the set $\Delta$ is the convex hull with vertexes the origin $\bm{O}$ and $\bm{w}_{1},\dots,\bm{w}_{n}$, i.e. a simplex.

We supplement the bases $\bm{m}_{1},\dots,\bm{m}_{t}$ of linear relations among roots by integral vectors $\bm{m}_{t+1},\dots,\bm{m}_{n}$ so that the $(n,n)$-matrix $M=(m_{i,j})$ with the $i$-th row $\bm{m}_{i}$ is in $SL_{n}(\mathbb{Z})$ and $\sigma(M):=(m_{i,\sigma^{-1}(j)})$, that is its $i$-th row is $\sigma(\bm{m}_{i})$. Then, writing

[TABLE]

we have

[TABLE]

and via the transformation $\bm{x}\mapsto\bm{y}:=\bm{x}{}^{t}\!(\sigma(M))$ we see that, noting $(\sigma(\bm{m}_{j}),\bm{x})=(\sigma(\bm{m}_{j})\sigma(M)^{-1},\bm{x}{}^{t}\!(\sigma(M)))=y_{j}$

[TABLE]

which is the parallel translation of the $(n-t)$-dimensional subspace $\{(0,\dots,0,\linebreak[3]y_{t+1},\dots,y_{n})\mid y_{i}\in\mathbb{R}\}$ by the vector $(k_{1},\dots,k_{t},0,\dots,0)$. The dimension of the set $\mathfrak{D}$ is at most $n-t$, which is the dimension of the supporting space $\mathcal{S}$.

Since the set $\Delta$ is the convex hull of points $\bm{O}$ and $\bm{w}_{1},\dots,\bm{w}_{n}$, we see that $\mathfrak{D}\,(=\mathfrak{D}(f,\sigma,\{k_{j}\}))$ is also a convex hull.

It is clear that vertexes of $\mathfrak{D}$ are in $\mathbb{Q}^{n}$ and the congruence $\bm{x}\equiv\bm{x}^{\prime}\bmod\mathbb{Z}^{n}$ for $\bm{x}\neq\bm{x}^{\prime}$ of $\mathfrak{D}(\subset\Delta\subset[0,1]^{n})$ implies that $\bm{x},\bm{x}^{\prime}$ are vertexes of $\Delta$. The vertex $\bm{w}_{l}$ is in $\mathfrak{D}$ if and only if $\sum_{i=l}^{n}m_{j,\sigma^{-1}(i)}=k_{j}$ $(1\leq{}^{\forall}j\leq t)$ holds.

Denote the $i$-th row of the matrix ${}^{t}\!(\sigma(M))^{-1}$ by $\bm{m}^{\prime}_{i}$, that is $(\sigma(\bm{m}_{j}),\bm{m}^{\prime}_{i})=\sigma(\bm{m}_{j}){}^{t}\bm{m}^{\prime}_{i}=\delta_{i,j}$, and put

[TABLE]

Lemma 6.1**.**

Write $\bm{r}:=(r_{1},\dots,r_{n})$ for local roots $r_{i}$ for $p\in Spl(f,\sigma,\{k_{j}\})$; then we have

[TABLE]

and

[TABLE]

holds except a finite number of primes under the assumption at the beginning of this subsection on the polynomial $f$.

Proof.

The first and second identities are obvious, and the third follows from $(\sigma(\bm{m}_{j}),\bm{r}/p)=k_{j}+m_{j}/p$ $(j=1,\dots,t)$ by definition. To see the last inclusion, we have only to show that the number of primes $p$ satisfying $(\bm{r}-\bm{c})/p\not\in\Delta$ is finite. If $(\bm{r}-\bm{c})/p\not\in\Delta$ happens for infinitely many primes, then one of conditions $0\leq r_{1}<c_{1}$, $0\leq r_{i+1}-r_{i}<c_{i+1}-c_{i}$ or $0<p-r_{n}<-c_{n}$ occurs for infinitely many primes, hence one of $r_{1}=c$, $r_{i+1}-r_{i}=c$ or $r_{n}-p=c$ for some constant $c$ occurs for infinitely many primes. This means $f(c)\equiv 0\bmod p$, $f(r_{i})\equiv f(r_{i}+c)\bmod p$, $f(c)\equiv 0\bmod p$, respectively. The first or the third condition implies easily that the polynomial $f(x)$ has a rational root $c$. The second implies that the polynomial $f$ is of the form $g(x)g(x-c)h(x)$. Because, if the identity $r_{i+1}-r_{i}=c$ holds for infinitely many primes, then $f(r_{i}+c)\equiv f(r_{i})\equiv 0\bmod p$ holds for infinitely many primes, which implies $(f(x),f(x+c))\neq 1$ (resp. $(f(x),f^{\prime}(x))\neq 1$) for $c\neq 0$ (resp. $c=0$), Hence, in the case of $c\neq 0$, there is a decomposition $f(x)=g(x)h(x),f(x+c)=g(x)k(x)$ for an irreducible polynomial $g(x)$, i.e. $f(x)$ is divisible by $g(x)g(x-c)$. If $c=0$, then $f(x)$ is of the form $g(x)^{2}h(x)$.

∎

Let $D$ be a subset of $\mathbb{R}^{n}$ satisfying conditions

[TABLE]

Take a point $\bm{x}$ in $D^{\circ}\cap\Delta^{\circ}\cap\mathcal{S}$ and suppose that a subsequence $\bm{r}/p\in D$ $(p\in Spl(f,\sigma,\{k_{j}\}))$ converges to $\bm{x}$. If $p$ is sufficiently large, then the point $(\bm{r}-\bm{c})/p\,(\in{\mathcal{S}})$ is close to $\bm{x}$, hence is in $D^{\circ}\cap\Delta^{\circ}\cap\mathcal{S}(\subset{D}\cap\mathfrak{D})$ if $p>n_{\bm{x}}$ for some number $n_{\bm{x}}$ depending on $\bm{x}$.

Neglecting the boundary, suppose the following equality

[TABLE]

which implies

[TABLE]

Now, under the assumption (74) the equation (13) in Conjecture 3 is equivalent to

[TABLE]

and the equation for the Weyl criterion is

[TABLE]

for a continuous function $F$ on $\mathfrak{D}$.

Last, let us discuss the estimate (74), i.e. whether the conjecture

[TABLE]

is true or not. May we expect the stronger estimate $O(1)$ than the $o$-estimate in the above?

Let us show this in the special case.

Proposition 6.1**.**

Let the set $D$ be of the form $D=\{\bm{x}\mid\sum_{j=1}^{n}d_{i,j}x_{j}\leq d_{i}\,(i=1,\dots,l)\}$ with all $d_{i,j},d_{i}$ being rational. If $\sum_{j=1}^{n}d_{i,\sigma(j)}\alpha_{j}\not\in\mathbb{Q}$ holds for $1\leq{}^{\forall}i\leq l$ and any permutation $\sigma$, then the equation (74) is true for the set $D$.

Proof.

Let $d$ be a positive integer such that all $dd_{i,j},dd_{i}$ are integers. Let us show the contradiction under the supposition that there are infinitely many primes $p\in Spl(f)$ such that $\bm{r}/p\in D$ and $(\bm{r}-\bm{c})/p\not\in D$; then we have $\sum_{j}d_{i,j}r_{j}\leq pd_{i}$ for all $i$ and $\sum_{j}d_{k,j}(r_{j}-c_{j})>pd_{k}$ for some $k$, hence $\sum_{j}dd_{k,j}c_{j}<\sum_{j}dd_{k,j}r_{j}-pdd_{k}\leq 0$, where $\sum_{j}dd_{k,j}c_{j}$ is a constant independent of $p$ and $\sum_{j}dd_{k,j}r_{j}-pdd_{k}$ is an integer. Therefore there are infinitely many primes $p\in Spl(f)$ such that $\sum_{j}dd_{k,j}r_{j}-pdd_{k}=\kappa$ for some integer $\kappa$ satisfying $\sum_{j}dd_{k,j}c_{j}<\kappa\leq 0$. Taking a prime ideal $\mathfrak{p}$ of $\mathbb{Q}(f)$ over $p$, we see that there are infinitely many prime ideals $\mathfrak{p}$ and a permutation $\sigma$ such that $\sum_{j}dd_{k,j}\sigma(\alpha_{j})\equiv\kappa\bmod\mathfrak{p}$, which implies $\sum_{j}dd_{k,j}\sigma(\alpha_{j})=\kappa$, i.e. $\sum_{j}dd_{k,\sigma^{-1}(j)}\alpha_{j}=\kappa$, which implies the contradiction $(d_{k,\sigma^{-1}(1)},\dots,d_{k,\sigma^{-1}(n)})\in LR_{0}$. The case of $\bm{r}/p\not\in D$ and $(\bm{r}-\bm{c})/p\in D$ is similarly proved. ∎

Corollary 6.2**.**

Suppose that the polynomial $f(x)$ has no rational root. Then (74) is true for the set of the form $D=\{\bm{x}\mid a_{i}\leq x_{i}\leq b_{i}\,(^{\forall}i)\}$ with all $a_{i},b_{i}$ being rational.

Proof.

Vectors $(d_{i,1},\dots,d_{i,n})$ in Proposition 6.1 are $(0,\dots,0,\pm 1,0,\dots,0)$. Such a vector is in $LR_{0}$ if and only if $f(x)$ has a rational root. ∎

How about the case of $d_{i,j}$ being irrational or more general boundaries?

6.3 Calculation of the integral

Let us recall

[TABLE]

and let us consider trigonometric functions on $\mathcal{S}$ as the function $F$ at (75). For $\bm{y}=\bm{x}{}^{t}\!(\sigma(M))$ and $\bm{a}=(0,\dots,0,a_{t+1},\dots,a_{n})$, we have

[TABLE]

hence the trigonometric function

[TABLE]

is basic on $\mathcal{S}$, and parameterizing $\mathcal{S}{}^{t}\!(\sigma(M))$ by $y_{t+1},\dots,y_{n}$

[TABLE]

where the projection $pr$ means

[TABLE]

Thus, we must ask whether the following is true or not :

[TABLE]

Here we assume $\#Spl(f,\sigma,\{k_{j}\})=\infty$, of course. We note that the condition $pr(\bm{x}{}^{t}(\sigma(M)))\equiv pr(\bm{y}{}^{t}(\sigma(M)))\bmod\mathbb{Z}^{n-t}$ for distinct points $\bm{x},\bm{y}\in\mathfrak{D}=\Delta\cap\mathcal{S}$ implies $\bm{x}-\bm{y}\in\mathbb{Z}^{n}$ and $\bm{x},\bm{y}\in\Delta$, i.e. $\bm{x},\bm{y}$ are vertexes of $\Delta$. By $T(\bm{x},\bm{a})=e(\sum_{i=t+1}^{n}a_{i}y_{i})$, functions $T(\bm{x},\bm{a})$ $(a_{t+1},\dots,a_{n}\in\mathbb{Z})$ separate two distinct points in $\mathfrak{D}$ except vertexes, hence under the assumption (74), the equation (77) holds for all integers $a_{t+1},\dots,a_{n}\linebreak[3]\in\mathbb{Z}$ if and only if (73) holds for every continuous function whose values are the same at non-separated vertexes.

A question is: Whether are all integral vectors $(0,\dots,0,a_{t+1},\dots,a_{n})$ really necessary to separate two distinct points in $\mathfrak{D}$ except vertexes or not? Let us see that if $\mathfrak{D}$ is an $(n-t)$-dimensional simplex, that is the convex hull of $n-t+1$ vertexes $\bm{v}_{1},\dots,\bm{v}_{n-t+1}$ such that $\bm{v}_{1}-\bm{v}_{n-t+1},\dots,\bm{v}_{n-t}-\bm{v}_{n-t+1}$ are linearly independent, then vectors $\bm{a}=(0,\dots,0,a_{t+1},\dots,a_{n})$ satisfying $(\bm{a},\bm{v}_{i})\in\mathbb{Z}$ $(i=1,\dots,n-t+1)$ are enough (cf. the remark at the end of this subsection).

For the convex hull $\mathcal{D}\,(\subset\mathbb{R}^{n})$ of points $\bm{v}_{1},\dots,\bm{v}_{m}$, i.e.

[TABLE]

it is clear

[TABLE]

A point $pr(\bm{v}_{i})$ is not necessarily a vertex of $pr(\mathcal{D})$ even if $\bm{v}_{i}$ is a vertex of $\mathcal{D}$.

To evaluate the integral (77), we divide $pr(\mathfrak{D}{}^{t}(\sigma(M)))$ to the union of simplexes and then apply Lemma 6.2 and Proposition 6.2 below. How can one divide and how can one write down the sum of integrals easily to see?

Let

[TABLE]

be an $m$-dimensional simplex, where $m+1$ vectors $\bm{v}_{1},\dots,\bm{v}_{m+1}$ are in $\mathbb{R}^{m}$ and $m$ vectors $\bm{v}_{1}-\bm{v}_{m+1},\dots,\bm{v}_{m}-\bm{v}_{m+1}$ are linearly independent. Then, putting

[TABLE]

and V=\left(\begin{array}[]{c}\bm{v}_{1}-\bm{v}_{m+1}\\ \vdots\\ \bm{v}_{m}-\bm{v}_{m+1}\end{array}\right), we have

[TABLE]

because putting $(y_{1},\dots,y_{m})=(x_{1},\dots,x_{m})V+\bm{v}_{m+1}$, we see that the integral is equal to

[TABLE]

and

[TABLE]

Here, we write $\bm{a}:=(a_{1},\dots,a_{m})$ and the second equality holds under the assumption that all $(\bm{v}_{i}-\bm{v}_{m+1},\bm{a})$ $(i=1,\dots,m+1)$ are mutually distinct (cf. Lemma 6.2 below). It is clear that

[TABLE]

Lemma 6.2**.**

Let $m$ be a positive integer. For real numbers $A_{1},\dots,A_{m}$ with $\prod_{i\neq j}(A_{i}(A_{i}-A_{j}))\neq 0$, we have

[TABLE]

which is equal, letting $A_{i}=\alpha_{i}-\alpha_{m+1}$ to

[TABLE]

and the integral vanishes if all $A_{i}$ are integers. Moreover, we have

[TABLE]

i.e.

[TABLE]

and for a polynomial $g(x)=c_{m-1}x^{m-1}+c_{m-2}x^{m-2}+\dots$

[TABLE]

Proof.

Denote the integral by $S(A_{1},\dots,A_{m})$. We use the induction on $m$. It is easy to see that $S(A_{1})=\frac{e(A_{1})-1}{2\pi iA_{1}}$ and $S(A_{1},\dots,A_{m})$ is equal to

[TABLE]

The assumption of the induction completes easily the proof of the first equality. W.r.t. the transform to $\alpha_{i}$, we have only to use the Vandermonde determinant:

[TABLE]

For the second identities, replace $A_{i}$ by $\epsilon A_{i}$; then we have

[TABLE]

which implies the second identities.

∎

Proposition 6.2**.**

Let $\alpha_{1},\dots\alpha_{m+1}\in\mathbb{R}$ and let $a_{1},\dots,a_{u}$ be their distinct numbers, i.e. $\{\alpha_{i}\mid 1\leq i\leq{m+1}\}=\{a_{i}\mid 1\leq i\leq u\}$ and $C_{J}:=\{j\mid\alpha_{j}=a_{J}\,(1\leq j\leq m+1)\}$. Then we have

[TABLE]

where the indexes $l,d$ run over $\{1,\dots,m+1\}\setminus C_{J}$.

Proof.

Let $\delta_{i}$ be mutually distinct numbers and put $\beta_{i}:=\alpha_{i}+\epsilon\delta_{i}$; then we see that $\beta_{i}-\beta_{j}=\alpha_{i}-\alpha_{j}+(\delta_{i}-\delta_{j})\epsilon$ is not [math] if $i\neq j$ and the non-zero number $\epsilon$ is sufficiently close to [math]. By Lemma 6.2, we have

[TABLE]

where

[TABLE]

and

[TABLE]

Hereafter the notations $\dot{\sum}_{k,l},\dot{\prod}_{k,l}$ mean the sum, the product on $k,l$ under the condition $1\leq k<l\leq m+1$. First, we see that

[TABLE]

where $\kappa:=\#\{(k,l)\mid 1\leq k<l\leq m+1,\alpha_{k}=\alpha_{l}\}$ and $O(\epsilon^{\kappa+1})$ means a power series whose coefficient of $\epsilon^{q}$ is [math] if $q<\kappa+1$.

Next, we see that

[TABLE]

where $\kappa_{j}:=\#\{(k,l)\mid 1\leq k<l\leq m+1,k,l\neq j,\alpha_{k}=\alpha_{l}\}$. Here we note that $\kappa_{j}$ depends only on $C_{J}$ containing $j$ by the equation $\kappa-\kappa_{j}=\#\{(k,l)\mid 1\leq k<l\leq m+1,\alpha_{k}=\alpha_{l},k\text{ or }l=j\}=\#C_{J}-1$. By

[TABLE]

we have

[TABLE]

Here, by the equality

[TABLE]

the product $\underset{k,l\neq j,\alpha_{k}\neq\alpha_{l}}{\dot{\prod}}(1+\frac{\delta_{k}-\delta_{l}}{\alpha_{k}-\alpha_{l}}\epsilon)$ is equal to

[TABLE]

hence

[TABLE]

The partial sum on $j\in C_{J}$ is

[TABLE]

applying (78) to the sum on $j\in C_{J}$ with $\delta_{j}-\delta_{m+1}$ for $A_{i}$ there, which is equal to

[TABLE]

Thus we have

[TABLE]

which completes the proof. ∎

Let us give a remark. Suppose that vertexes $\bm{v}_{1},\dots,\bm{v}_{m+1}$ are in $\mathbb{Q}^{m}$ such that $\bm{v}_{1}-\bm{v}_{m+1},\dots,\bm{v}_{m}-\bm{v}_{m+1}$ are linearly independent. Denote by $V$ the regular matrix whose $i$th row is $\bm{v}_{i}-\bm{v}_{m+1}$ as before, and put

[TABLE]

For vectors $\bm{a}\in\mathcal{A}_{0}$, the condition $\prod_{i\neq j}A_{i}(A_{i}-A_{j})\neq 0$ just before Lemma 6.2 is satisfied. Let the vector $\bm{a}_{j}$ be the $j$th row of the matrix ${}^{t}V^{-1}$, i.e. $(\bm{v}_{i}-\bm{v}_{m+1},\bm{a}_{j})=\delta_{i,j}$, hence $\bm{a}_{j}\in\mathcal{A}$ is clear. Since the module $\mathcal{A}$ is spanned by vectors $\bm{a}_{j}$, the module $U$ is the the dual of $\mathcal{A}$, hence we see the equivalence

[TABLE]

Let us see that the module $\mathcal{A}$ is spanned by $\mathcal{A}_{0}$. Take a vector $\bm{a}$ and write $\bm{b}:=\bm{a}+\sum_{l=1}^{m}k_{l}\bm{a}_{l}$ $(k_{i}\in\mathbb{Z})$. The conditions $(\bm{b},\bm{v}_{i}-\bm{v}_{j})\neq 0$ and $(\sum_{l=1}^{m}k_{l}\bm{a}_{l},\bm{v}_{i}-\bm{v}_{j})\neq 0$ are equivalent to $(\bm{a},\bm{v}_{i}-\bm{v}_{j})+\sum k_{l}(\delta_{i,l}-\delta_{j,l})\neq 0$ and $\sum k_{l}(\delta_{i,l}-\delta_{j,l})\neq 0$, that is $(\bm{a},\bm{v}_{i}-\bm{v}_{j})+k_{i}-k_{j}\neq 0$ and $k_{i}-k_{j}\neq 0$. The existence of such integers $k_{l}$ is easily proved by the induction on $m$. Therefore the equation $\bm{a}=\bm{b}-\sum_{l=1}^{m}k_{l}\bm{a}_{l}$ implies that $\mathcal{A}$ is spanned by $\mathcal{A}_{0}$

Assume that $\bm{x},\bm{y}\in\mathcal{D}$ satisfies $e((\bm{x},\bm{a}))=e((\bm{y},\bm{a}))$, i.e, $(\bm{x}-\bm{y},\bm{a})\in\mathbb{Z}$ for all $\bm{a}\in\mathcal{A}_{0}$, hence for all $\bm{a}\in\mathcal{A}$ as above; then we have $\bm{x}-\bm{y}\in U$. Writing $\bm{x}=\sum_{i=1}^{m}x_{i}(\bm{v}_{i}-\bm{v}_{m+1})+\bm{v}_{m+1},\bm{y}=\sum_{i=1}^{m}y_{i}(\bm{v}_{i}-\bm{v}_{m+1})+\bm{v}_{m+1}$ $(0\leq{}^{\forall}x_{i},{}^{\forall}y_{i},\sum_{j=1}^{m}x_{j},\sum_{j=1}^{m}y_{j}\leq 1)$, we have $\bm{x}-\bm{y}=\sum_{i=1}^{m}(x_{i}-y_{i})(\bm{v}_{i}-\bm{v}_{m+1})\in U$. Suppose $x_{j}-y_{j}=1$ for some $j$; then we have $x_{j}=1$, i.e. $\bm{x}=\bm{v}_{j}$ and $\bm{x}-\bm{y}=(1-y_{j})(\bm{v}_{j}-\bm{v}_{m+1})+\sum_{i\neq j}(-y_{i})(\bm{v}_{i}-\bm{v}_{m+1})\in U$, and so ${}^{\forall}y_{i}=0,1$, hence $\bm{y}=\bm{v}_{k}$ for some $k$. Thus we see that $\bm{x},\bm{y}$ are vertexes or $\bm{x}=\bm{y}$ if $\bm{x},\bm{y}\in\mathcal{D}$ satisfies $e((\bm{x},\bm{a}))=e((\bm{y},\bm{a}))$ for all $\bm{a}\in\mathcal{A}_{0}$. In other words, the functions $e((\bm{x},\bm{a}))$ $(\bm{a}\in\mathcal{A}_{0})$ can separate distinct points in $\mathcal{D}$ except vertexes.

6.4 Example

First, we note that for an integer $k$ $(1\leq k\leq n-1)$, vertexes of

[TABLE]

are following $k(n-k)+1$ vectors:

[TABLE]

with non-negative integers $p,q,r$ satisfying $p+q+r=n$ and

[TABLE]

Because, if $\bm{x}\in\Delta_{k}$ satisfies three strict inequalities $x_{i}<x_{i+1}=\dots=x_{i+a}<x_{i+a+1}=\dots=x_{i+a+b}<x_{i+a+b+1}$, then for any $\eta$ sufficiently close to [math], the vector $\bm{x}(\eta)$, replacing $x_{i+1}=\dots=x_{i+a},x_{i+a+1}=\dots=x_{i+a+b}$ by $x_{i+a}+\eta/a,x_{i+a+1}-\eta/b$, respectively is still in $\Delta_{k}$, that is a short segment containing $\bm{x}$ is in $\Delta_{k}$.

For $1\leq k\leq n-1$, the point $P(0,n,0)=(\frac{k}{n},\dots,\frac{k}{n})$ is the vertex, but points $P(n,0,0),P(0,0,n)$ are not vertexes by virtue of $\sum x_{i}=0,n$, respectively. The case $q=1$ does not occur, and for $2\leq q\leq n$, $r=\min(k-1,n-q),p=n-q-r$ gives a vertex.

Example

Vertexes of $\Delta_{1}$ are $(\underbrace{0,\dots,0}_{n-q},\underbrace{\frac{1}{q},\dots,\frac{1}{q}}_{q})$ $(1\leq q\leq n)$.

Vertexes of $\Delta_{n-1}$ are $(0,1,\dots,1)$ and $(\underbrace{\frac{q-1}{q},\dots,\frac{q-1}{q}}_{q},\underbrace{1,\dots,1}_{n-q})$ $(2\leq q\leq n)$.

In general, a point $\bm{x}\in\mathfrak{D}(f,\sigma,\{k_{j}\})$ is not a vertex, that is there is a non-zero vector $\bm{v}=(v_{1},\dots,v_{n})$ such that $\bm{x}+\eta\bm{v}\in\mathfrak{D}(f,\sigma,\{k_{j}\})$ holds for any $\eta$ in a short open interval containing [math] if and only if there is a non-zero vector $\bm{v}$ satisfying that

• •

$\sum_{i}m_{j,\sigma^{-1}(i)}v_{i}=0$ for ${}^{\forall}j=1,\dots,t$, and

• •

$v_{i}=0,v_{i}=v_{i+1},v_{i}=0$ hold according to $x_{i}=0,x_{i}=x_{i+1},x_{i}=1$, respectively.

(1) Let a polynomial $f(x)$ have no non-trivial linear relation; then we can take vectors $\bm{m}_{1},\dots,\bm{m}_{n}$ in the previous subsection as

[TABLE]

which are rows of the unimodular matrix $M$ as before. By $\mathfrak{D}=\mathfrak{D}(f,\sigma,k)=\Delta_{k}$, $pr(\mathfrak{D}{}^{t}\!M)(\subset\mathbb{R}^{n-1})$ is the convex hull of vectors whose $i$-th entry $(i=1,\dots,n-1)$ is the $(i+1)$-th one of $P(p,q,r)$.

• •

In the case of $k=1$ : Vertexes of $pr(\mathfrak{D}{}^{t}\!M)(\subset\mathbb{R}^{n-1})$ are

[TABLE]

where the number $\frac{1}{n-i+1}$ begins from $(i-1)$-th entry, and the $(n-1,n-1)$-matrix $V$ with the $i$th row $\bm{v}_{i}-\bm{v}_{n}$ $(i=1,\dots,n-1)$ is

[TABLE]

and

[TABLE]

hence the dual module $\mathcal{A}$ of the module spanned by rows of $\bm{v}_{i}-\bm{v}_{n}$ is spanned by rows of ${}^{t}V^{-1}$, i.e.

[TABLE]

The index $[\mathbb{Z}^{n-1}:\mathcal{A}]$ is equal to $n!$ by $\det(V^{-1})=(-1)^{n-1}n!$. In this case, $pr(\mathfrak{D}{}^{t}\!M)$ is a simplex.

• •

In the case of $k=n-1$ : Vertexes of $pr(\mathfrak{D}{}^{t}\!M)(\subset\mathbb{R}^{n-1})$ are $\bm{v}_{1}:=(1,\dots,1),\bm{v}_{q}:=(\underbrace{\frac{q-1}{q},\dots,\frac{q-1}{q}}_{q-1},\underbrace{1,\dots,1}_{n-q})$ $(q=2,\dots,n)$, hence the inclusion

[TABLE]

is clear.

• •

In the case of $(n,k)=(4,2)$ : Vertexes of $pr(\mathfrak{D}{}^{t}\!M)$ are

[TABLE]

and writing $C_{i}:=\bm{v}_{i+1}-\bm{v}_{1}$ $(i=1,2,3,4)$, i.e.

[TABLE]

we see that the linear relation $C_{3}+C_{4}=\frac{4}{3}C_{1}+\frac{2}{3}C_{2}$ and $\{C_{1},C_{2},C_{3}\}$ and $\{C_{1},C_{2},C_{4}\}$ are linearly independent, respectively. By the equation

[TABLE]

the domain $pr(\mathfrak{D}{}^{t}\!M)$ is a union of two $3$-dimensional simplexes as

[TABLE]

Let V_{1}=\left(\begin{array}[]{c}C_{1}\\ C_{2}\\ C_{3}\end{array}\right),V_{2}=\left(\begin{array}[]{c}C_{1}\\ C_{2}\\ C_{4}\end{array}\right); then we see

[TABLE]

Hence the dual module $\mathcal{A}$ of the module spanned by the rows of $V_{i}$ is spanned by columns of $V_{i}^{-1}$, hence $\mathcal{A}\subset\mathbb{Z}^{3}$. It is easy to see that $|V_{1}|=\frac{-1}{12},|V_{2}|=\frac{1}{12}$, hence their absolute values are equal. Thus we see that

[TABLE]

where $u_{i}:=(\bm{v}_{i},(a_{1},a_{2},a_{3}))$, $e_{i}:=e(u_{i})$ and all $u_{i}$ are supposed to be distinct. There are relations $u_{4}=\frac{4}{3}u_{2}-\frac{2}{3}u_{3}$, $u_{5}=\frac{4}{3}u_{3}$ by $\bm{v}_{4}=\frac{4}{3}\bm{v}_{2}-\frac{2}{3}\bm{v}_{3}$, $\bm{v}_{5}=\frac{4}{3}\bm{v}_{3}$, and $\mathbb{Z}[\bm{v}_{1},\dots,\bm{v}_{5}]=\mathbb{Z}[\bm{v}_{1},3\bm{v}_{2}-2\bm{v}_{4}-\bm{v}_{5},\bm{v}_{3}-\bm{v}_{5}]$ with $3\bm{v}_{2}-2\bm{v}_{4}-\bm{v}_{5}=\frac{1}{3}\bm{v}_{2},\bm{v}_{3}-\bm{v}_{5}=\frac{-1}{3}\bm{v}_{3}$.

We get the same division to the union of the simplexes for the base vertex $\bm{v}_{2}$ instead of $\bm{v}_{1}$, but for other vertex $\bm{v}_{i}$ with $i\neq 1,2$ one of the corresponding dual modules is not integral, for example $pr(\mathfrak{D}{}^{t}\!M)$ is a union of two simplexes as

[TABLE]

where simplexes correspond to $x_{2}\geq 2x_{3}$, $x_{2}\leq 2x_{3}$ in order, and for U_{1}:=\left(\begin{array}[]{c}\bm{v}_{1}-\bm{v}_{5}\\ \bm{v}_{4}-\bm{v}_{5}\\ \bm{v}_{2}-\bm{v}_{5}\end{array}\right),U_{2}:=\left(\begin{array}[]{c}\bm{v}_{1}-\bm{v}_{5}\\ \bm{v}_{4}-\bm{v}_{5}\\ \bm{v}_{3}-\bm{v}_{5}\end{array}\right) we see

[TABLE]

The determinants of $U_{1},U_{2}$ are $\frac{1}{18},\frac{-1}{9}$, respectively, hence their absolute values are distinct.

• •

In the case of $1\leq k\leq n-1$ : Vertexes of $\Delta_{k}$ are $P(p,q,r)$ defined above : If $r=k$, then

[TABLE]

If $0\leq r\leq k-1,(k-r)+1\leq q\leq(k-r)+(n-k)=n-r$, then

[TABLE]

Therefore we see : If $r=k$, then

[TABLE]

If $0\leq r\leq k-1,q=n-r$, then

[TABLE]

If $0\leq r\leq k-1,(k-r)+1\leq q<n-r$, then

[TABLE]

What is the concrete division of $pr(\mathfrak{D}{}^{t}\!M)$ to simplexes and what is the integral of $e(\sum_{i=1}^{n-1}a_{i}y_{i})$ on it ?

(2) Next, let us consider the case of the polynomial $f(x)$ of degree $4$ which has a non-trivial linear relation among roots. By Proposition 3.3, we may assume that their basis are $\alpha_{1}+\alpha_{4},\alpha_{2}+\alpha_{3}\in\mathbb{Z}$, hence

[TABLE]

and $\bm{m}_{1}=(1,0,0,1),\bm{m}_{2}=(0,1,1,0),\bm{m}_{3}=(0,0,1,0),\bm{m}_{4}=(0,0,0,1)$. Hence we see

[TABLE]

for which vertexes of the simplex $pr(\mathfrak{D}{}^{t}\!M)$ are $(\frac{1}{2},\frac{1}{2}),(\frac{1}{2},1),(1,1)$, and for V=\left(\begin{array}[]{cc}\frac{-1}{2}&\frac{-1}{2}\\ \frac{-1}{2}&0\end{array}\right), V^{-1}=\left(\begin{array}[]{cc}0&-2\\ -2&2\end{array}\right) is integral.

7 Numerical data

In this section, the integer $X_{m}=X_{m}(f)$ denotes the smallest prime number $p\in Spl(f)$ with $p>10^{m}$.

7.1 Case without non-trivial linear relation among roots

Let $f(x)$ be a polynomial of degree $n$ without non-trivial linear relation among roots. First we check (14), i.e. (11) for a special domain $D_{i}=\{(x_{1},\dots,x_{n})\in[0,1)^{n}\mid x_{i}\leq a\}$ at $a=k/(10n)$. “diff” in the table means the maximum of the absolute value of the difference between $1$ and the left-hand side divided by the right-hand side in (14) at $a=k/(10n)$ $(k=1,\dots,10n)$, $i=1,\dots,n$ and $X=X_{m}$ for $m=5,\dots,11$. “Diff” means the maximum of the absolute value of the difference

[TABLE]

for $\{R_{i}\}\in\mathfrak{R}(f,\sigma,k,L)$.

(1) $f(x)=x^{2}+1$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

(2) $f(x)=x^{3}+2$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Hereafter, we give cases $(1\leq k\leq\deg f-1,2\leq L\leq 8)$ when $\text{Diff}>0.1$ at $m=9$. Comparing with $\#\mathfrak{R}(f,\sigma,k,L)(\geq L^{n-1})$, $m=9$ seems to be small.

(3) $f(x)=x^{4}+x^{3}+x^{2}+x+1$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

(4) $f(x)=x^{5}+2$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

For the following, $m=9$ is too small to get $\text{Diff}<0.1$.

(5) $f(x)=x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$

[TABLE]

(6) $f(x)=x^{7}+2$

[TABLE]

7.2 Case of degree $6$ with non-trivial linear relation among roots

Let $f(x)=x^{6}+14x^{4}+49x^{2}+7$, which is the fourth polynomial in Example 3.1. Then taking roots there, we see that $t=4$ and the basis of linear relations

[TABLE]

Let us determine permutations $\sigma$ satisfying $\dim\mathfrak{D}(f,\sigma)=2$. To do it, suppose that $0<y_{1}<\dots<y_{6}<1,y_{1}+y_{6}=y_{2}+y_{5}=y_{3}+y_{4}=1$ and the corresponding equation to $\hat{\bm{m}_{4}}$

[TABLE]

Then possible combinations above are $y_{1}+y_{2}+y_{3}=1$ or $y_{1}+y_{2}+y_{4}=1$. Identifying a permutation $\sigma\in S_{6}$ with the image $[\sigma(1),\dots,\sigma(6)]$, the set of representatives of cosets of $\sigma$ with $\dim\mathfrak{D}(f,\sigma)=2$ by the group $\hat{\bm{G}}$ is $\sigma_{1}=[1,2,4,6,5,3],\sigma_{2}=[1,2,3,6,5,4]$, supposing that $\sigma(1)=1,\sigma(4)=6,\sigma(2)=2,\sigma(5)=5,\{\sigma(3),\sigma(6)\}=\{3,4\}$ by $\{\{\sigma(1),\sigma(4)\}$, $\{\sigma(2),\sigma(5)\}$, $\{\sigma(3),\sigma(6)\}\}=$ $\{\{1,6\}$, $\{2,5\}$, \{3,4\}$$\}.

[TABLE]

Since the length of $\bm{u}_{i},\bm{v}_{i}$ $(i=1,2)$ is $2$ and the angle of $\bm{u}_{1},\bm{u}_{2}$ (resp. $\bm{v}_{1},\bm{v}_{2}$) is $\pi/3$ and the volumes of their projection to $(x_{1},x_{2})$-space are $1/16,1/48$, we see $vol(\mathfrak{D}(f,\sigma_{1}))=3vol(\mathfrak{D}(f,\sigma_{2}))$. Therefore Conjecture 1 is numerically confirmed as follows : In the following, “diff1” (resp. “diff2”) means the difference $\#Spl_{X_{m}}(f,\sigma_{1})/\#Spl_{X_{m}}(f)-3/4$ (resp. $\#Spl_{X_{m}}(f,\sigma_{2})/\#Spl_{X_{m}}(f)-1/4$).

[TABLE]

Let us check Conjecture 4. We assume that $\sigma=\sigma_{1}$ or $\sigma_{2}$ above and $k_{1}=\dots=k_{4}=1$ because of $\#Spl_{\infty}(f,\sigma,\{k_{j}\})<\infty$ otherwise. Denoting $\#\mathfrak{R}(f,\sigma,\{k_{j}\},L)$ by $R$, and by “er” the maximum of

[TABLE]

with $\{R_{i}\}\in\mathfrak{R}(f,\sigma,\{k_{j}\},L)$, we see

$\sigma=\sigma_{1},L=2,(R=4)$

[TABLE]

$\sigma=\sigma_{1},L=3,(R=144)$

[TABLE]

$\sigma=\sigma_{1},L=4,(R=256)$

[TABLE]

$\sigma=\sigma_{1},L=5,(R=6400)$

[TABLE]

$\sigma=\sigma_{1},L=6,(R=576)$

[TABLE]

$\sigma=\sigma_{1},L=7,(R=63504)$

[TABLE]

$\sigma=\sigma_{2},L=2,(R=4)$

[TABLE]

$\sigma=\sigma_{2},L=3,(R=144)$

[TABLE]

$\sigma=\sigma_{2},L=4,(R=256)$

[TABLE]

$\sigma=\sigma_{2},L=5,(R=6400)$

[TABLE]

$\sigma=\sigma_{2},L=6,(R=576)$

[TABLE]

$\sigma=\sigma_{2},L=7,(R=63504)$

[TABLE]

7.3 Case of degree $4$ with non-trivial linear relation among roots

Let us consider the polynomial $f(x)=x^{4}+1$ in this subsection. First we check (11) numerically for $D_{i}=\{(x_{1},\dots,x_{4})\mid x_{i}\leq A\}$. To do it, as in the subsection 4.2 we have only to check

[TABLE]

and the right-hand side is given by $8vol(pr(D_{i}\cap\mathfrak{D}(f,id))$ as in that subsection.

“diff” in the table means the maximum of the absolute value of the difference between $1$ and the left-hand side divided by the right-hand side at $A=k/40$ $(k=1,\dots,40)$, $i=1,\dots,4$ and $X=X_{m}$ for $m=5,\dots,11$.

[TABLE]

Next, let us check Conjecture 4. As in the subsection 4.2 again, we have only to consider the case of $\{\sigma(1),\sigma(4)\}=\{1,4\}$ or $\{2,3\}$ and $k_{1}=k_{2}=1$. Then

[TABLE]

Therefore, we see

[TABLE]

The following are tables of the maximum of

[TABLE]

at $\{R_{i}\}\in\mathfrak{R}(f,\sigma,\{k_{j}\},L)$ for $m=5,\dots,10$ and $L=2,\dots,10$.

$L=2$

[TABLE]

$L=3$

[TABLE]

$L=4$

[TABLE]

$L=5$

[TABLE]

$L=6$

[TABLE]

$L=7$

[TABLE]

$L=8$

[TABLE]

$L=9$

[TABLE]

$L=10$

[TABLE]

7.4 Reducible polynomial

7.4.1

Let us consider the polynomial $f(x)=(x^{2}+x+1)(x^{2}+2x+2)$ with roots $\alpha_{i}$ such that $\alpha_{i}^{2}+\alpha_{i}+1=0$ for $i=2,3$ and $\alpha_{i}^{2}+2\alpha_{i}+2=0$ for $i=1,4$. The field $\mathbb{Q}(f)$ is $\mathbb{Q}(\sqrt{-3},\sqrt{-1})$. Vectors $\hat{\bm{m}}_{i}$ be

[TABLE]

Then we see that

[TABLE]

and

[TABLE]

hence $\dim\mathfrak{D}(f,\sigma)=2\,(=n-t)$ if and only if $\sigma\in\bm{G}$. For $X=10^{8}$, we find

[TABLE]

Three conditions $\sigma\in\bm{G}$, $Pr(f,\sigma)>0$ and $vol(\mathfrak{D}(f,\sigma))>0$ are equivalent.

For the one-dimensional uniformity, we see that, with $X=10^{8}$

[TABLE]

With respect to Conjecture 2, let $D$ be $\{(x_{1},x_{2},x_{3},x_{4})\mid x_{1}<1/3\}$. The absolute value of the difference between

[TABLE]

for $X=10^{8}$ and $\sigma\in\bm{G}$ is $0.0003$. With respect to Conjecture 4, we see $k_{1}=k_{2}=1$ and $\bm{G}/\hat{\bm{G}}=\{[1,2,3,4],[2,1,4,3]\}$. In the following table, the maximum of errors

[TABLE]

over $\{R_{i}\}\in\mathfrak{R}(f,\sigma,\{k_{j}\},L)$ is given for $\sigma\in\bm{G}/\hat{\bm{G}}$, and $\#\mathfrak{R}$ is $\#\mathfrak{R}(f,\sigma,\{k_{j}\},L)$, which is independent of $\sigma$.

[TABLE]

7.4.2

Let us consider the polynomial $f(x)=(x^{2}-2)((x-1)^{2}-2)$ with roots $\alpha_{1}=\sqrt{2},\alpha_{2}=1+\sqrt{2},\alpha_{3}=-\sqrt{2},\alpha_{4}=1-\sqrt{2}$. Then the matrix $(m_{j,i})$ and $m_{i}$ $(t=3)$ are

[TABLE]

and $\bm{G}\!=\!\{[1,2,3,4],[1,2,4,3],[2,1,3,4],[2,1,4,3],[3,4,1,2],[3,4,2,1],[4,3,1,2]$, $[4,3,2,1]\}$ and $\hat{\bm{G}}=\{[1,2,3,4],[3,4,1,2]\}$, where a permutation $\sigma$ is identified with the vector $[\sigma(1),\dots,\sigma(4)]$ of images. Then we see that

[TABLE]

so two conditions $\dim\mathfrak{D}(f,\sigma)=1\,(=n-t)$ and $\sigma\in G$ are equivalent. We find

[TABLE]

since $p\in Spl(f,\sigma)$ means $r_{\sigma(1)}+r_{\sigma(4)}=p+1,r_{\sigma(1)}-r_{\sigma(2)}=-1,r_{\sigma(1)}+r_{\sigma(3)}=p$ for a sufficiently large $p$, which implies $\sigma\in\hat{\bm{G}}$. This is an example such that $Pr(f,\sigma)=0$ but $vol(\mathfrak{D}(f,\sigma))>0$ for $\sigma\in\bm{G}\setminus\hat{\bm{G}}$ by $\#\bm{G}=8,\#\hat{\bm{G}}=2$, and this supports Conjecture 1. We see easily that for $\sigma\in\bm{G}$

[TABLE]

For the one-dimensional uniformity, we see that, with $X=10^{8}$

[TABLE]

The one-dimensional uniformity follows surely from [DFI], since $f(x)$ is a product of quadratic polynomials which define the same quadratic field $\mathbb{Q}(\sqrt{2})$.

With respect to Conjecture 2, let $D$ be $\{(x_{1},x_{2},x_{3},x_{4})\mid x_{1}<1/3\}$. The absolute value of the difference between

[TABLE]

for $X=10^{8}$ and $\sigma=id$ is $0.00032$. To check Conjecture 4, we see that $k_{1}=1,k_{2}=0,k_{3}=1$, $\sigma\in\hat{\bm{G}}$ hold if $Pr(f,\sigma,\{k_{j}\},L,\{R_{i}\})>0$, and for them the maximum $er$ of errors

[TABLE]

over $\{R_{i}\}\in\mathfrak{R}(f,\sigma,\{k_{j}\},L)$ is as follows :

[TABLE]

7.4.3

Let us consider the polynomial $f(x)=(x^{2}+1)(x^{2}-2)(x^{4}-2x^{2}+9)$, whose roots are $\pm\sqrt{2},\pm\sqrt{-1},\pm\sqrt{2}\pm\sqrt{-1}$ and we number them as $\alpha_{1}=\sqrt{-1}$, $\alpha_{2}=\sqrt{2}-\sqrt{-1}$, $\alpha_{3}=\sqrt{2}$, $\alpha_{4}=-\sqrt{2}-\sqrt{-1}$, $\alpha_{5}=\sqrt{2}+\sqrt{-1}$, $\alpha_{6}=-\sqrt{2}$, $\alpha_{7}=-\sqrt{2}+\sqrt{-1}$, $\alpha_{8}=-\sqrt{-1}$. The rows of the following matrix $M$ gives $\bm{m}_{j}$ $(j=1,\dots,6)$ with $m_{1}=\dots=m_{6}=0$, i.e. $M\cdot{}^{t}(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5},\alpha_{6},\alpha_{7},\alpha_{8})=0$ $(t=6)$:

[TABLE]

The group $\bm{G}$ is, by Proposition 4.1

[TABLE]

where a permutation $\sigma$ is identified with the vector $[\sigma(1),\dots,\sigma(8)]$ of its images. The set of $\sigma$ satisfying (6) for some $p$ less than $10^{8}$ consists of $\sigma_{i}G$ for

[TABLE]

The corresponding domain $\mathfrak{D}(f,\sigma_{i})=\{\bm{x}=C+x_{1}\bm{v}_{1}+x_{2}\bm{v}_{2}\mid S\}$ are given by the following: Vectors $C,\bm{v}_{i}$ are

[TABLE]

and for $i=7$,

[TABLE]

The inequalities $S$ on $x_{1},x_{2}$ are

[TABLE]

The volumes $[vol(\mathfrak{D}(f,\sigma_{1})),\dots,vol(\mathfrak{D}(f,\sigma_{7}))]$ are $[1/4,3/40,1/20,3/40,1/8,\linebreak 1/20,1/8]$ with $1/4+\dots+1/8=3/4$. Let us give a brief proof on $\sigma_{1}$. Since $(\bm{v}_{1},\bm{v}_{1})=12,(\bm{v}_{1},\bm{v}_{2})=6,(\bm{v}_{2},\bm{v}_{2})=6$, we define the orthonormal vectors $\bm{u}_{i}$ by $\bm{v}_{1}=2\sqrt{3}\bm{u}_{1},\bm{v}_{2}=\sqrt{3}(\bm{u}_{1}+\bm{u}_{2})$. Writing $x_{1}\bm{v}_{1}+x_{2}\bm{v}_{2}=y_{1}\bm{u}_{1}+y_{2}\bm{u}_{2}$, we see $y_{1}=2\sqrt{3}x_{1}+\sqrt{3},y_{2}=\sqrt{3}x_{2}$ with Jacobian $6$. The volume of the set of $(x_{1},x_{2})$ defined by the condition $S$, i.e. $0\leq x_{1}\leq x_{2}\leq 1,4x_{1}+2x_{2}\leq 1$ is $1/24$, hence $vol(\mathfrak{D}(f,\sigma_{1}))=6\cdot 1/24=1/4$.

For $X=10^{8}$, the following is a table of $\#Spl_{X}(f,\sigma_{i})/\#Spl_{X}(f)$ on the first line and $vol(\mathfrak{D}(f,\sigma_{i}))/\sum_{i}vol(\mathfrak{D}(f,\sigma_{i}))$ on the second in the order of $i=1,\dots,7$, respectively

[TABLE]

Exact values on the second line are $[1/3,1/10,1/15,1/10,1/6,1/15,1/6]$ which are equal to $(3/4)^{-1}[1/4,3/40,1/20,3/40,1/8,1/20,1/8]$. This supports Conjecture 1.

For $p\leq 10^{8}$, integers $k_{j}$ in (7) are $k_{1}=\dots=k_{4}=1,k_{5}=k_{6}=0$ for all $\sigma_{i}$, and for such integers $k_{j}$, $D:=\{(x_{1},\dots,x_{8})\mid|\sum_{i}m_{j,i}x_{\sigma(i)}-k_{j}|<\epsilon\,(1\leq{}^{\forall}j\leq 6)\}$ $(\epsilon>0)$ contains $\mathfrak{D}(f,\sigma)$ for $\sigma=\sigma_{1},\dots,\sigma_{7}$, hence Conjectures 2 is true for such domains.

For the one-dimensional uniformity, we see that, with $X=10^{8}$

[TABLE]

For $L=2$ and $\{k_{j}\}$ above, $\mathfrak{R}(f,\sigma_{i},\{k_{j}\},2)$ with $\#\mathfrak{R}(f,\sigma_{i},\{k_{j}\},2)=4$ $(i=1,\dots,7)$ is in order

[TABLE]

where $\{R_{i}\}$ is written as $[R_{1},\dots,R_{8}]$.

For $X=10^{8}$, $\#Spl_{X}(f,\sigma_{i},\{k_{j}\},2,\{R_{i}\})/\#Spl_{X}(f,\sigma_{i},\{k_{j}\})$ is, corresponding to the above table

[TABLE]

The top left value $0.25030$ corresponds to $i=1$ and $\{R_{i}\}=[R_{1},\dots,R_{8}]=[0,0,0,0,1,1,1,1]$, and so on. All values are almost $1/4=0.25$ and this matches Conjecture 4.

8 $M(f,\mu)$ and the decimal part $\left\{\frac{g(r)}{p}\right\}$ again

Let us give observations on $M(f,\mu)$ defined in the subsection 4.1 and problems derived from them. Let us recall definitions and properties

[TABLE]

where $A\fallingdotseq B$ means that sets $A\setminus B,B\setminus A$ are finite. We note that in case of $Gal(\mathbb{Q}(f)/\mathbb{Q})\cong S_{n}$, $\hat{G}=G_{0}=S_{n}$ follows , hence the set $M(f,\mu)$ is equal to $Spl(f)$.

$\bullet$ Let us give an example of $M(f,\mu)\,(\neq Spl(f))$. Let $f(x)=x^{3}-3x+1$ which is abelian, and let roots be $\alpha_{1}=\alpha,\alpha_{2}=-\alpha^{2}-\alpha+2,\alpha_{3}=\alpha^{2}-2$. Then we see that $\hat{\bm{G}}=S_{3}$ and $\bm{G}_{0}=\{[1,2,3],[3,1,2],[2,3,1]\}$ where a permutation $\sigma$ is identified with $[\sigma(1),\sigma(2),\sigma(3)]$. Thus we see that $Spl(f)=Spl(f,\sigma)=\cup M(f,\sigma)=M(f,id)\cup M(f,(2,3))$ for the transposition $(2,3)\not\in\bm{G}_{0}$. Since for a prime $p\in M(f,id)$, we have $r_{2}\equiv\alpha_{2}=-\alpha^{2}-\alpha+2\equiv-r_{1}^{2}-r_{1}+2\mod\mathfrak{p}$, primes $p\in Spl(f)$ are separated as follows:

For $\mu=id.$, $M(f,\mu)=\{p\in Spl(f)\mid r_{2}\equiv-r_{1}^{2}-r_{1}+2\bmod p\}=\{37,73,89,181,233,251,269,397,449,467,521,541,557,593,613,631,683,809,\linebreak 811,919,\dots\}$.

For $\mu=(2,3)$, $M(f,\mu)=\{p\in Spl(f)\mid r_{2}\equiv r_{1}^{2}-2\bmod p\}=\{17,19,53,71,\linebreak 107,109,127,163,179,197,199,271,307,359,379,431,433,487,503,523,577,\linebreak 647,701,719,739,757,773,827,829,863,881,883,\dots\}$.

Their densities in $Spl(f)$ seem to be $1/2=1/[\hat{\bm{G}}:\bm{G}_{0}]$. The density of $M(f,\mu)$ in $Spl(f,\mu)$ is conjectured to be $\frac{[\mathbb{Q}(f):\mathbb{Q}]}{\#\hat{G}}$ for every permutation $\mu$ satisfying $\#M(f,\mu)=\infty$ just before Corollary 4.2.

$\bullet$ We introduce the equivalence relation $\sim$ among polynomials as follows : For monic integral polynomials $f(x),g(x)$ of degree $n$,

$f(x)\sim g(x)$ if and only if $g(x)=\delta^{n}f(\delta x+m)$

with $\delta=\pm 1$ and $m\in\mathbb{Z}$ holds. Denote the roots of $f(x)$ by $\alpha_{i}$ $(i=1,\dots,n)$ and assume that any root $\alpha_{i}$ is not a rational integer. Then $\beta_{i}:=\delta(\alpha_{i}-m)$ are roots of $g(x)$, and for local roots $r_{i}$ of $f(x)$ at $p\in Spl(f)=Spl(g)$ $r^{\prime}_{i}:=\delta(r_{i}-m)$ are roots of $g(x)\equiv 0\bmod p$, and $0<r_{1}-m\leq\dots\leq r_{n}-m<p$ if $p$ is sufficiently large. Hence writing $R_{i}:=r^{\prime}_{i}$ in the case of $\delta=1$, otherwise $R_{i}:=r^{\prime}_{n-i+1}$, we see that $R_{i}$ $(i=1,\dots,n)$ are local roots of $g(x)$ if $p$ is sufficiently large. Then we see that neglecting small primes

[TABLE]

where $\mu^{\prime}(i):=n+1-\mu(i)$. Thus the set of $M(f,\mu)$ $(\mu\in S_{n})$ depends on the equivalence class of polynomials. In case of $f(x)=x^{3}-3x+1,x^{4}+2x^{3}+3x^{2}-3x+1,x^{4}+3x^{2}-2x+2,x^{4}-x^{3}+2x^{2}+x+1,x^{4}+x^{3}+x^{2}+x+1,(x^{2})^{2}+x^{2}-1,(x^{2})^{2}+1,x^{4}+4x^{3}+2x^{2}-4x-1=(x^{2}+2x)^{2}-2(x^{2}+2x)-1$ which exhaust all types (with respect to $Gal(\mathbb{Q}(f)/\mathbb{Q})$ and the existence of non-trivial linear relations among roots) of polynomials of degree 3,4 with $[\hat{\bm{G}}:\bm{G}_{0}]>1$, it is likely that sets $M(f,\mu)$ $(\mu\in S_{n})$ with $\#M(f,\mu)=\infty$ characterize the above equivalence class among polynomials $g(x)$ under assumptions that $\mathbb{Q}(g)=\mathbb{Q}(f)$, $\deg g=\deg f$ and $f,g$ have the same linear relations among roots. When does the set of $M(f,\mu)$ $(\mu\in S_{n})$ characterize the equivalence class of polynomials above ?

$\bullet$ About the computer experiment on the question before Lemma 4.1 : In this paragraph, set $f(x)=x^{4}+1$ whose roots are $\alpha_{1}=\frac{1+\sqrt{-1}}{\sqrt{2}},\alpha_{2}=\frac{1-\sqrt{-1}}{\sqrt{2}}=-\alpha_{1}^{3},\alpha_{3}=-\frac{1-\sqrt{-1}}{\sqrt{2}}=\alpha_{1}^{3},\alpha_{4}=-\frac{1+\sqrt{-1}}{\sqrt{2}}=-\alpha_{1}$, where linear relations among roots over rationals are spanned by $\alpha_{1}+\alpha_{4}=\alpha_{2}+\alpha_{3}=0$. It is easy to see that $\#\hat{\bm{G}}=8,\#\bm{G}_{0}=4$ and the transposition $(3,4)$ is in $\hat{\bm{G}}\setminus\bm{G}_{0}$, and $\mathfrak{D}(f,\sigma)=\{(x_{1},\dots,x_{4})\mid 0\leq x_{1}\leq\dots\leq x_{4}\leq 1,x_{1}+x_{4}=x_{2}+x_{3}=1\}$ except finite points for $\sigma\in\hat{\bm{G}}$, which is identified to $\{(x_{1},x_{2})\mid 0\leq x_{1}\leq x_{2}\leq\frac{1}{2}\}$ and $\{(x_{3},x_{4})\mid\frac{1}{2}\leq x_{3}\leq x_{4}\leq 1\}$ by projections $(x_{1},\dots,x_{4})\mapsto(x_{1},x_{2}),(x_{3},x_{4})$, respectively.

For at most quadratic polynomials $g_{1}(x),g_{2}(x)\in\mathbb{Z}[x]$, let us consider

[TABLE]

where $r_{i}$’s are local roots of the polynomial $f(x)$ for the prime $p$ and $\{a\}$ denotes the decimal part of $a$, and write $M(X):=(M_{1}(X),\dots,M_{4}(X))$ and $\#M(X):=(\#M_{1}(X),\dots,\#M_{4}(X))$ simply. For non-zero integers $m,n$, the set $I_{m,n}$ of $x\in(0,1)$ satisfying $\{mx\}<\{nx\}$ is a union of intervals. Hence, for linear polynomials $g_{1}(x)=mx,g_{2}(x)=nx$ we see

[TABLE]

and the density $\lim_{X\to\infty}\frac{\#M_{i}(X)}{\#Spl_{X}(f)}$ is described by the domain of $\{\bm{x}\in(0,1)^{4}\mid x_{i}\in I_{m,n}\}$ by Conjecture. For example, for polynomials $g_{1}(x)=nx,g_{2}(x)=2nx$ $(n\neq 0,n\in\mathbb{Z})$, the computer experiment suggests

[TABLE]

which is equal to

[TABLE]

where

[TABLE]

and the denominator $\frac{1}{8}$ is the volume of the whole spaces $\{(x_{1},x_{2})\mid 0\leq x_{1}\leq x_{2}\leq\frac{1}{2}\}$, $\{(x_{3},x_{4})\mid\frac{1}{2}\leq x_{3}\leq x_{4}\leq 1\}$, which are identified with $\{(x_{1},\dots,x_{4})\mid 0\leq x_{1}\leq\dots\leq x_{4}\leq 1,x_{1}+x_{4}=x_{2}+x_{3}=1\}$ as above. These match the experiment above quite well and Conjecture 2.

Other cases, i.e. cases of $\deg(g_{1}(x))$ or $\deg(g_{2}(x))=2$ are under the assumption that $\deg(g_{1}(x))\leq 2,\deg(g_{2}(x))\leq 2,g_{1}(0)\linebreak[3]=g_{2}(0)=0$, (i) $\deg(g_{1}(x))=2$ and $g_{2}(x)=nx\,(n\neq 0)$, (ii) $\deg(g_{1}(x))=\deg(g_{2}(x))=2$ and $g_{1}(x)-g_{2}(x)=nx,(n\neq 0)$, or (iii) $\deg(g_{1}(x))=\deg(g_{2}(x))\linebreak[3]=\deg(g_{1}(x)-g_{2}(x))=2$, and in case of (i),(ii) the density looks like

[TABLE]

and in case of (iii) it looks like $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$. The author does not know how to elucidate this, in particular even the reason of

[TABLE]

How about the case of polynomials of degree $3$ ? Since we have $r_{3}\equiv r_{1}^{3}\bmod p$ for $p\in M(f,id)$ as in Example 2 below, the inequality $\{\frac{g_{1}(r_{1})}{p}\}<\{\frac{g_{2}(r_{1})}{p}\}$ is equivalent to $\{\frac{r_{3}}{p}\}<\{\frac{r_{1}^{2}}{p}\}$ for $g_{1}(x)=x^{3},g_{2}(x)=x^{2}$, for example. The density of such primes in $M(f,id.)$ looks like $\frac{1}{3}$.

It may be appropriate to start by $Spl(f,\sigma)$ or $M(f,\sigma)$ instead of $Spl(f)$ for a general polynomial.

$\bullet$ (cf. Problem on p.4.1) Let us consider the density of primes $p\in Spl(f)$ satisfying

[TABLE]

In case of $\deg g_{1}=\deg g_{2}=1$, it seems to be independent of the polynomial $f(x)$ with $t=1$ (cf. Subsection 8.1). But for polynomials $g_{1}(x)=x^{2},g_{2}(x)=-2x$ (resp. $g_{1}(x)=x^{2},g_{2}(x)=2x$), the density seems $\frac{7}{8},\frac{7}{8}-\frac{1}{16},\frac{7}{8}+\frac{1}{16}$ (resp. $\frac{7}{8},\frac{7}{8}+\frac{1}{16},\frac{7}{8}$ ) according as $f(x)=x^{3}-2,x^{3}+x^{2}-2x-1,x^{3}+3x^{2}-1$, where the latter two polynomials are abelian. How does the density depend on the polynomial $f$ ?

$\bullet$ Let $f(x)$ be an irreducible polynomial of degree $n$ with root $\alpha$, and suppose that $\mathbb{Q}(\alpha)$ is a Galois extension over $\mathbb{Q}$ with Galois group $\{\sigma_{1},\dots,\sigma_{n}\}$. Write

[TABLE]

Lemma 8.1**.**

Let a prime $p\in Spl(f)$ be sufficiently large and $r$ a root of $f(x)\equiv 0\bmod p$. Then all roots of $f(x)\equiv 0\bmod p$ are given by $c_{i}(r)\bmod p$ $(i=1,\dots,n)$.

Proof.

Let $\mathfrak{p}=(\alpha-r,p)$ be a prime ideal of $\mathbb{Q}(\alpha)$ over $p$ and let $\iota$ be the canonical homomorphism from the ring $\mathfrak{o}$ of integers of $\mathbb{Q}(\alpha)$ on $\mathfrak{o}/\mathfrak{p}\cong\mathbb{Z}/p\mathbb{Z}$. Then $\iota(\sigma_{i}(\alpha))$ $(i=1,\dots,n)$ are all roots of $f(x)\equiv 0\bmod p$, and so $r\bmod p=\iota(\sigma_{t}(\alpha))$ for some $t$. Hence we see that $\{\iota(\sigma_{j}(\alpha))\mid j=1,\dots,n\}=\{\iota(\sigma_{t}(\sigma_{i}(\alpha)))\mid i=1,\dots,n\}=\{\iota(c_{i}(\sigma_{t}(\alpha))\mid i=1,\dots,n\}=\{c_{i}(r)\mid i=1,\dots,n\}$.

∎

Let us consider the density of the complement set, that is the problem that for not necessarily monic polynomials $g_{1}(x),g_{2}(x)$ over $\mathbb{Z}$ what the density of primes $p\in Spl(f)$ satisfying

[TABLE]

is. By Lemma 8.1 the above inequality is equivalent to

[TABLE]

for some root $r$ of $f(x)\equiv 0\bmod p$ except finitely many primes. Defining rational numbers $g_{l,i,k}$ by

[TABLE]

we expect that the density of primes $p\in Spl(f)$ satisfying

[TABLE]

is equal to the volume of the set

[TABLE]

Here the reason why the term for $k=0$ is omitted is due to Proposition 8.1 below. The computer experiment supports the above in the case that $f(x)=x^{3}-3x+1$ and $g_{1},g_{2}$ are polynomials such that degrees are less than or equal to $2$ and the absolute values of their coefficients are less than $4$.

In case that the polynomial $f(x)$ is not Galois, the author has no perspective. For example, in case of $f(x)=x^{3}-2,x^{3}+x+1$, the density seems to be $\frac{7}{8}$ independently of $g_{1}(x),g_{2}(x)$. How do we elucidate it?

$\bullet$ The above expectation suggests that the sequence of vectors

[TABLE]

is uniformly distributed in $[0,1)^{n-1}$ where $r$ runs $n$ roots of $f(x)\bmod p$ with $p\in Spl(f)$. Applying it to the domain $\{\bm{x}\in[0,1)^{n-1}\mid a\leq x_{k}\leq b\}$, it implies the uniform distribution of the sequence $\left\{\frac{r^{k}}{p}\right\}$ $(p\in Spl(f))$ where $r$ runs $n$ distinct roots of $f(x)\equiv 0\bmod p$ and $k$ is a fixed integer with $1\leq k\leq n-1$. We can not replace $Spl(f)$ by $M(f,\sigma)$.

$\bullet$ Next, let us consider polynomials in many variables : Write

[TABLE]

Let $g(\bm{x}),h(\bm{x})\in\mathbb{Z}[x_{1},\dots,x_{n}]$. For a permutation $\mu$, what is the density

[TABLE]

Since an element in $\mathbb{Q}(f)$ is a polynomial in $\alpha_{1},\dots,\alpha_{n}$, there is a polynomial $g_{1}(\bm{x})\in\mathbb{Q}(f)[\bm{x}]$ such that $g(\alpha_{1},\dots,\alpha_{n})=g_{1}(\alpha_{1},\dots,\alpha_{n})$. Thus, we see that $g(r_{\mu(1)},\dots,r_{\mu(n)})\equiv g_{1}(r_{\mu(1)},\dots,r_{\mu(n)})\bmod p$ for a prime $p\in M(f,\mu)$. In general, $g(r_{1},\dots,r_{n})\equiv g_{1}(r_{1},\dots,r_{n})\bmod p$ $({}^{\forall}p\in Spl(f))$ is not necessarily true. If $g_{1}(\bm{x})$ and similarly $h_{1}(\bm{x})$ are linear with integer coefficients, then we have only to refer to Conjecture $2$ if $Spl(f,\mu)=M(f,\mu)$. A sufficient condition to $g_{1}(\bm{x})$ being linear is that $\mathbb{Q}(f)/\mathbb{Q}$ is a Galois extension and $t=1$, because they imply that $[\mathbb{Q}(f):\mathbb{Q}]=n$ and $\dim\mathbb{Q}[\alpha_{1},\dots,\alpha_{n},1]=n$, that is $\mathbb{Q}(f)=\mathbb{Q}[\alpha_{1},\dots,\alpha_{n},1]$. Otherwise, what happens? Is the density the volume of some figure? We give below two examples $f(x)=x^{3}+2$, which is non-Galois, and $f(x)=x^{4}+1$ in the case of $t=2>1$.

Example 1 Let $f(x)=x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\in\mathbb{Q}[x]$ be an irreducible polynomial with roots $\alpha_{1},\alpha_{2},\alpha_{3}$ and suppose that the extension $\mathbb{Q}(\alpha_{1})/\mathbb{Q}$ is not a Galois extension. Then for a polynomial $g(\bm{x})\in\mathbb{Q}[x_{1},x_{2},x_{3}]$, there are rational numbers $c_{i}$ such that $g(\alpha_{1},\alpha_{2},\alpha_{3})=c_{0}+c_{1}\alpha_{1}+c_{2}\alpha_{1}^{2}+c_{3}\alpha_{2}+c_{4}\alpha_{1}\alpha_{2}+c_{5}\alpha_{1}^{2}\alpha_{2}$. Note that $\alpha_{1}\alpha_{2}$ may be replaced by $\alpha_{2}^{2}$ by the equation $\alpha_{1}\alpha_{2}=-a_{1}-a_{2}(\alpha_{1}+\alpha_{2})-\alpha_{1}^{2}-\alpha_{2}^{2}$.

Let us give computer experiments with respect to the above density $D_{f}(g,h)\linebreak[3]=D_{f}(g,h;id)$ for $f(x)=x^{3}+2$. Polynomials $g(\bm{x}),h(\bm{x})$ are restricted to be of the following special type without the term $x_{1}^{2}x_{2}$ for simplicity

[TABLE]

with rational integer coefficients $g_{i}$ without constant term by Proposition 8.1. Denote its homogeneous part of degree $i$ by $g_{i}(\bm{x})$. For polynomials $g(\bm{x}),h(\bm{x})$ with $g(\bm{x})\neq 0,h(\bm{x})\neq 0,g(\bm{x})\neq h(\bm{x})$ as above, let us give observations. If both of $g,h$ are linear, i.e. $g(\bm{x})=g_{1}(\bm{x}),h(\bm{x})=h_{1}(\bm{x})$, then we have only to refer to Conjecture 2, and so we suppose that $g(\bm{x})$ or $h(\bm{x})$ is not linear.

Lemma 8.3 implies

[TABLE] 2. 2.

The case of $g(\bm{x})=g_{1}(\bm{x})$ and $h_{2}(\bm{x})\neq 0$.

[TABLE] 3. 3.

The case of $g(\bm{x})=g_{2}(\bm{x})$ and $h(\bm{x})=h_{2}(\bm{x})$. The density $D_{f}(g,h)$ is always $\frac{1}{2}$. 4. 4.

The case of $g(\bm{x})=g_{2}(\bm{x})$, $h_{1}(\bm{x})\neq 0$ and $h_{2}(\bm{x})\neq 0$. The density $D_{f}(g,h)$ is

[TABLE] 5. 5.

The case of $g_{2}(\bm{x})\neq 0,g_{1}(\bm{x})\neq 0,h_{2}(\bm{x})\neq 0,h_{1}(\bm{x})\neq 0.$ The density $D_{f}(g,h)$ is, writing $k_{i}:=g_{i}-h_{i}$

[TABLE]

These seem to be true for other non-Galois irreducible polynomials of degree $3$.

Example 2 Set $f(x)=x^{4}+1$ and let roots be $\alpha_{1}:=\frac{1+\sqrt{-1}}{\sqrt{2}},\alpha_{2}:=\frac{1-\sqrt{-1}}{\sqrt{2}}=-\alpha_{1}^{3},\alpha_{3}:=-\frac{1-\sqrt{-1}}{\sqrt{2}}=\alpha_{1}^{3},\alpha_{4}:=-\frac{1+\sqrt{-1}}{\sqrt{2}}=-\alpha_{1}$ as before, and then the basis $\bm{m}_{j}$ of $LR_{0}\cap\mathbb{Z}^{4}$ are $(1,0,0,1),(0,1,1,0)$. Noting that $\mathbb{Q}(f)=\mathbb{Q}[1,\alpha_{1},\alpha_{1}^{2},\alpha_{2}]$, we see that for a polynomial $g(\bm{x})$ in $\mathbb{Q}[x_{1},\dots,x_{4}]$, there are rational numbers $c_{i}$ so that $g(\alpha_{1},\dots,\alpha_{4})=c_{0}+c_{1}\alpha_{1}+c_{2}\alpha_{1}^{2}+c_{3}\alpha_{2}$. So, we may assume that $g(\bm{x})=c_{0}+c_{1}x_{1}+c_{2}x_{1}^{2}+c_{3}x_{2}$. For a permutation $\mu$, let us determine $M(f,\mu)$. Since the condition $p\in M(f,\mu)$ is equivalent to $\alpha_{i}\equiv r_{\mu(i)}\bmod\mathfrak{p}$, for $\sigma\in Gal(\mathbb{Q}(f)/\mathbb{Q})$ satisfying $\sigma(\alpha_{\mu^{-1}(1)})=\alpha_{1}$, it is equivalent to $\sigma(\alpha_{i})\equiv r_{\mu(i)}\bmod\sigma(\mathfrak{p})$, i.e. $\sigma(\alpha_{\mu^{-1}(i)})\equiv r_{i}\bmod\sigma(\mathfrak{p})$, hence we have $M(f,\mu)=M(f,\mu^{\prime})$ with $\mu^{\prime}(1)=1$ for $\mu^{\prime}$ defined by $\alpha_{\mu^{\prime}(i)}=\sigma(\alpha_{\mu^{-1}(i)})$. Hence we assume $\mu(1)=1$. For a prime $p\in M(f,\mu)$, the conditions $\alpha_{1}+\alpha_{4}=0,r_{1}+r_{4}=p$ imply $\mu(4)=4$, hence $\mu(3)=2$ or $3$. By $\alpha_{3}=\alpha_{1}^{3}\equiv r_{1}^{3}\bmod\mathfrak{p}$, we have $r_{\mu(3)}\equiv r_{1}^{3}\bmod p$. Thus we have, supposing $\mu(1)=1$ $\mu=id.$ or the transposition $(2,3)$ and

[TABLE]

Note that $Spl(f)=Spl(f,id.)=M(f,id.)\cup M(f,(2,3))$. Under the assumptions

the densities of $M(f,id.)$ and $M(f,(2,3))$ in $Spl(f)$ are equal to $\frac{1}{2}$, 2. 2.

the densities $D_{f}(g,h;id.),D_{f}(g,h;(2,3))$ are equal,

we see that $D_{f}(g,h;\mu)=\frac{1}{2}(D_{f}(g,h;id.)+D_{f}(g,h;(2,3)))$ is equal to

[TABLE]

Thus, in case that $g(\bm{x}),h(\bm{x})$ are linear, the density is given by Conjecture 2. For example, let $g_{1}(\bm{x})=2x_{1}-x_{2},g_{2}(\bm{x})=x_{2}$; then the computer experiment suggests $D_{f}(g_{1},g_{2};\mu)=\frac{1}{2}$, which match $vol(\{(x_{1},x_{2})\mid 0<x_{1}<x_{2}<\frac{1}{2},\{2x_{1}-x_{2}\}<\{x_{2}\}\})=vol(\{(x_{1},x_{2})\mid 0<x_{1}<x_{2}<\frac{1}{2},x_{1}<x_{2}<2x_{x}\})=\frac{1}{16}$ and $vol(\{(x_{1},x_{2})\mid 0<x_{1}<x_{2}<\frac{1}{2}\})=\frac{1}{8}$.

Let us give other observations : Write

[TABLE]

and suppose that $g(\bm{x})\neq 0,h(\bm{x})\neq 0,g(\bm{x})\neq h(\bm{x})$. Then the observation is that, independently of $\mu$ $D_{f}(g,h):=D_{f}(g,h;\mu)$ $(\mu=id.$ or $(2,3))$ is

[TABLE] 2. 2.

The case of $g(\bm{x})=g_{2}x_{1}^{2},h(\bm{x})=h_{2}x_{1}^{2}$.

[TABLE] 3. 3.

The case of $g(\bm{x})=g_{2}x_{1}^{2},h(\bm{x})=h_{1}x_{1}+h_{3}x_{2}$.

[TABLE] 4. 4.

The case of $g(\bm{x})=g_{2}x_{1}^{2},h(\bm{x})=h_{1}x_{1}+h_{2}x_{1}^{2}+h_{3}x_{2}$ with $h_{2}\neq 0$, and $h_{1}\neq 0$ or $h_{3}\neq 0$. Then $D_{f}(g,h)$ is

[TABLE] 5. 5.

The case of $g(\bm{x})=g_{1}x_{1}+g_{3}x_{2},h(\bm{x})=h_{1}x_{1}+h_{2}x_{1}^{2}+h_{3}x_{2}$ with $h_{2}\neq 0$, and $h_{1}\neq 0$ or $h_{3}\neq 0$. Then $D_{f}(g,h)$ is

[TABLE] 6. 6.

The case of $g(\bm{x})=g_{1}x_{1}+g_{2}x_{1}^{2}+g_{3}x_{2},h(\bm{x})=h_{1}x_{1}+h_{2}x_{1}^{2}+h_{3}x_{2}$ with $g_{2}h_{2}\neq 0$, and $g_{1}\neq 0$ or $g_{3}\neq 0$, and $h_{1}\neq 0$ or $h_{3}\neq 0$. Then $D_{f}(g,h)$ is, writing $k_{i}:=g_{i}-h_{i}$

[TABLE]

$\bullet$ In this paragraph, for integers $m,p$, we denote the integer $a$ satisfying $a\equiv m\bmod p$ and $0\leq a<p$ by $[m]_{p}$, and let the polynomial $f(x)$ be irreducible and of degree $n\,(>1)$ with roots $\alpha_{i}$ as usual.

Lemma 8.2**.**

Let $a$ be a positive integer.

(i)* For a polynomial $g(x)\in\mathbb{Z}[x]$, there are only finitely many primes $p$ satisfies $[g(r)]_{p}=a$ (resp. $[g(r)]_{p}=p-a$) in the case of $g(x)-a$ ( resp. $g(x)+a$) being not divisible by $f(x)$ in the ring $\mathbb{Q}[x]$.*

(ii)* For a polynomial $g(\bm{x})\in\mathbb{Z}[x_{1},\dots,x_{n}]$, there are only finitely many primes $p\in Spl(f)$ such that $[g(\bm{r})]_{p}=a$ (resp. $[g(r)]_{p}=p-a$) if $g(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})\neq a$ (resp. $g(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})\neq-a$) for any permutation $\sigma$. Here $\bm{r}=(r_{1},\dots,r_{n})$ is the vector of local roots for $p\in Spl(f)$.*

Proof.

(i) Suppose that there are infinitely many primes $p$ satisfying $[g(r)]_{p}=a$. Writing $g_{0}(x)=g(x)-a$, we see that there are infinitely many primes $p$ such that there is an integer $r$ satisfying $f(r)\equiv g_{0}(r)\equiv 0\bmod p$. Since the polynomial $f(x)$ is irreducible and $g_{0}(x)$ is not divisible by $f(x)$, there are integral polynomials $h_{1}(x),h_{2}(x)$ and a non-zero integer $h$ such that $h_{1}(x)f(x)+h_{2}(x)g_{0}(x)=h$. So $h$ is divisible by infinitely many primes as above, which is the contradiction. In the other case, we have only to put $g_{0}(x)=g(x)+a$.

(ii) Suppose that there are infinitely many primes $p\in Spl(f)$ such that $[g(\bm{r})]_{p}=a$. Then there is a permutation $\sigma$ such that $\alpha_{\sigma(i)}\equiv r_{i}\bmod\mathfrak{p}$ for infinitely many primes $p\in Spl(f)$, where $\mathfrak{p}$ is a prime ideal over $p$ in $\mathbb{Q}(f)$, which implies $g(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})\equiv a\mod\mathfrak{p}$, that is $g(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})=a$. It is the contradiction. The proof of the other case is similar. ∎

Lemma 8.3**.**

Let $a$ be a positive integer. (i) Let $g_{i}(x)$ $(i=1,2)$ be integral polynomials for which $g_{2}(x)-g_{1}(x)-a$ is not divisible by $f(x)$. Then there are only finitely many primes $p\in Spl(f)$ such that

[TABLE]

(ii)* Let $g_{i}(\bm{x})$ $(i=1,2)$ be polynomials in $\mathbb{Z}[x_{1},\dots,x_{n}]$ and $a$ a positive integer. Suppose that $g_{2}(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})-g_{1}(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})\neq a$ for every permutation $\sigma$. Then there are only finitely many primes $p\in Spl(f)$ such that*

[TABLE]

Proof.

(i) Suppose that there are infinitely many primes $p\in Spl(f)$, which satisfies the equality in question. By writing $g(x):=g_{2}(x)-g_{1}(x)$, the supposition means that there are infinitely many primes $p$ such that $[g(r)]_{p}\equiv[g_{2}(r)]_{p}-[g_{1}(r)]_{p}=a\bmod p$, hence $[g(r)]_{p}=a$. The previous lemma completes the proof.

(ii) Assume that there are infinitely many primes $p$ satisfying $[g_{2}(\bm{r})]_{p}-[g_{1}(\bm{r})]_{p}=a$. Writing $g(\bm{x}):=g_{2}(\bm{x})-g_{1}(\bm{x})$, we see that $[g(\bm{r})]_{p}\equiv[g_{2}(\bm{r})]_{p}-[g_{1}(\bm{r})]_{p}\equiv a\bmod p$, hence $[g(\bm{r})]_{p}=a$. The previous lemma completes the proof.

∎

Proposition 8.1**.**

Let $a,b$ be integers.

(i)* Let $g_{i}(x)$ $(i=1,2)$ be polynomials over $\mathbb{Z}$ such that none of $g_{1}(x),g_{2}(x)$, $g_{2}(x)-g_{1}(x)$ are congruent to a constant modulo $f(x)$. Then for a prime $p$ with an integer $r$ satisfying $f(r)\equiv 0\bmod p$, the inequalities $\{\frac{g_{1}(r)+a}{p}\}<\{\frac{g_{2}(r)+b}{p}\}$ and $\{\frac{g_{1}(r)}{p}\}<\{\frac{g_{2}(r)}{p}\}$ are equivalent except a finitely many primes.*

(ii)* Let $g_{i}(\bm{x})$ $(i=1,2)$ be polynomials in $\mathbb{Z}[x_{1},\dots,x_{n}]$ such that none of $g_{1}(\bm{x}),g_{2}(\bm{x})$, $g_{2}(\bm{x})-g_{1}(\bm{x})$ are rational integers at $\bm{x}=(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})$ for every permutation $\sigma$. Then for a prime $p\in Spl(f)$, the inequalities $\{\frac{g_{1}(\bm{r})+a}{p}\}<\{\frac{g_{2}(\bm{r})+b}{p}\}$ and $\{\frac{g_{1}(\bm{r})}{p}\}<\{\frac{g_{2}(\bm{r})}{p}\}$ are equivalent except a finitely many primes.*

Proof.

(i) For simplicity, the notation $A\overset{f}{\Leftrightarrow}B$ means that statements $A$ and $B$ are equivalent except finitely many primes $p$ in this proof. We see that

[TABLE]

which is equivalent to $[g_{1}(r)]_{p}<[g_{2}(r)]_{p}$, i.e. $\{\frac{g_{1}(r)}{p}\}<\{\frac{g_{2}(r)}{p}\}$ in the case of $b-a\geq 0$. Suppose $b-a<0$; then the above inequality is $[g_{1}(r)]_{p}<[g_{2}(r)]_{p}+b-a$. Lemma 8.3 says that there are only finitely many primes $p$ satisfying $[g_{2}(r)]_{p}+b-a\leq[g_{1}(r)]_{p}\leq[g_{2}(r)]_{p}$, hence we have $[g_{1}(r)]_{p}<\min([g_{2}(r)]_{p},[g_{2}(r)]_{p}+b-a)\overset{f}{\Leftrightarrow}[g_{1}(r)]_{p}<[g_{2}(r)]_{p}$. Thus, we have $\{\frac{g_{1}(r)+a}{p}\}<\{\frac{g_{2}(r)+b}{p}\}\overset{f}{\Leftrightarrow}[g_{1}(r)]_{p}<[g_{2}(r)]_{p}\,(\Leftrightarrow\{\frac{g_{1}(r)}{p}\}<\{\frac{g_{2}(r)}{p}\})$. The other case is similar. ∎

8.1 Polynomial without non-trivial linear relations among roots

In this subsection, assume that the polynomial $f(x)$ is of degree $\tilde{n}\,(>1)$ and has no non-trivial linear relations among roots, i.e. $t:=\dim LR=1$, and for linear forms $g_{1}(x)=mx,g_{2}(x)=nx$ with distinct non-zero integers $m,n$, we study the density

[TABLE]

where $\{\frac{g_{i}(r)}{p}\}$ denotes the decimal part of $\frac{g_{i}(r)}{p}$ as usual.

The condition $t=1$ implies $\hat{\bm{G}}=S_{n}$, and so $Spl(f,{}^{\forall}\sigma)=Spl(f)$. We see that

[TABLE]

hence

[TABLE]

since the number of the prime $p$ satisfying $\{\frac{nr}{p}\}=\{\frac{mr}{p}\}$ is finite and we write

[TABLE]

Let us evaluate the ratio of volumes. Note that the dimension $\dim I_{n,m}^{\tilde{n}}\cap\hat{\mathfrak{D}}_{\tilde{n}}$ is less than or equal to ${\tilde{n}}-1$ and the volume is considered as the $({\tilde{n}}-1)$-dimensional set under the assumption $\dim LR=1$. We know $vol(\hat{\mathfrak{D}}_{\tilde{n}})=\frac{\sqrt{{\tilde{n}}}}{{\tilde{n}}!}$. On the other hand, we see that

[TABLE]

and the permutation induces the orthogonal transformation on $\mathbb{R}^{\tilde{n}}$, and the dimension of the intersection of the subsets corresponding to distinct permutation is less than ${\tilde{n}}-1$, hence we see

[TABLE]

and

[TABLE]

In the rest of this subsection we will show the following :

Theorem 8.1**.**

For integers $M,N$ satisfying $0<N<M$ and $(M,N)=1$, we have

[TABLE]

where $B_{\tilde{n}}(x)$ is the Bernoulli polynomial.

The above matches computer experiment of $d_{M,N}$. We note that

• (i)

the ratio of volumes is independent of $M,N$ for odd $\tilde{n}>1$, since $B_{\tilde{n}}(0)=0$ for odd $\tilde{n}>1$.

• (ii)

Let $f(x)$ be quadratic; then the above shows $\frac{vol(I_{N,M}^{2}\cap\hat{\mathfrak{D}}_{2})}{vol(\hat{\mathfrak{D}}_{2})}=0$. As a matter of fact, we show, more strongly in Proposition 8.4 that $I_{N,M}^{2}\cap\hat{\mathfrak{D}}_{2}$ is a finite set. The set $\{p\in Spl_{X}(f)\mid\{\frac{nr}{p}\}<\{\frac{mr}{p}\}\text{ for }^{\forall}r\text{ s.t. }f(r)\equiv 0\bmod p\}$ is also a finite one as in Proposition 8.6.

8.1.1 Bernoulli polynomial

In this subsection, let us recall several facts on Bernoulli polynomials. For non-negative integer $m$, we write

[TABLE]

for example

[TABLE]

We define Bernoulli polynomials $B_{n}(t)$ by

[TABLE]

that is

[TABLE]

In this paper, the notation $B_{n}$ denotes the Bernoulli polynomial, and so we use the notation $B_{n}(0)$ for the Bernoulli number.

We see that, for a real number $a$ and a non-negative integer $m$

[TABLE]

hence for integers $n,m\geq 0$,

[TABLE]

Therefore, for a positive real number $a$ and a positive integer $n>1$, we see that

[TABLE]

which we need to prove the theorem.

8.1.2 Proof of Theorem 8.1

It is convenient to introduce the notation $\fallingdotseq$ : For sets $S,T$, we write $S\fallingdotseq T$ if and only if $\#(S\setminus T)+\#(T\setminus S)<\infty$.

Proposition 8.2**.**

Let $n,m$ be non-zero distinct integers, and write $n=dN$, $m=dM$ for $d:=(n,m)$. Then we see that

[TABLE]

and under the assumptions $M>N>0$ and $(M,N)=1$

[TABLE]

where the integer $L$ is defined by

[TABLE]

Proof.

The condition $x\in I_{n,m}$ means $\{nx\}<\{mx\}$, i.e. $\{dNx\}<\{dMx\}$, hence $\{dx\}\in I_{N,M}$. We have only to write $dx=k+\{dx\}$ for an integer $k$ for the proof of (81). Let us show the equation (82). Take a number $x\in I_{N,M}$, which implies $Mx\not\in\mathbb{Z}$ and write $x=\frac{K-\epsilon}{M}$ $(1\leq K\leq M,K\in\mathbb{Z},0<\epsilon<1)$. We note that the condition $K=M$ is equivalent to $L=M$ by the definition of $L$ and the assumption $M>N,(M,N)=1$. Then it is easy to see that $\{Mx\}=1-\epsilon$ and $\{Nx\}=\{\frac{NK-N\epsilon}{M}\}=\{\frac{L-N\epsilon}{M}\}$. The inequalities $0<L-N\epsilon<2M$ show that $\{Nx\}=\frac{L-N\epsilon}{M}$ or $\frac{L-N\epsilon}{M}-1$ according to $L-N\epsilon<M$ or $L-N\epsilon\geq M$. Let us note that

(i) $\frac{L-M}{N}<\frac{2M-L}{M-N}$ which follows from the assumption $L<M+N$,

(ii) $\frac{L-N\epsilon}{M}-1<1-\epsilon\Leftrightarrow\epsilon<\frac{2M-L}{M-N}$, and

(iii) $\frac{L-N\epsilon}{M}<1-\epsilon\Leftrightarrow\epsilon<\frac{M-L}{M-N}$.

First, suppose $L<M$ : By $0<L-N\epsilon<M$, we have $\{Nx\}=\frac{L-N\epsilon}{M}$, hence $\{Nx\}<\{Mx\}$ $\Leftrightarrow$ $\frac{L-N\epsilon}{M}<1-\epsilon$ $\Leftrightarrow$ $0<\epsilon<\frac{M-L}{M-N}$ by (iii). Next, suppose $L\geq M$: Under the supposition $\{Nx\}<\{Mx\}$, the condition $L-N\epsilon<M$ implies $\{Mx\}=1-\epsilon>\{Nx\}=\frac{L-N\epsilon}{M}$, which is equivalent to $(0<)\,\epsilon<\frac{M-L}{M-N}$, which contradicts the assumption $L\geq M$. Hence the condition $\{Nx\}<\{Mx\}$ implies $L-N\epsilon\geq M$, i.e. $\epsilon\leq\frac{L-M}{N}$. Conversely, suppose $\epsilon\leq\frac{L-M}{N}$, which is equivalent to $L-N\epsilon\geq M$ and implies $\epsilon<\frac{2M-L}{M-N}$ by (i) , hence $\{Nx\}=\frac{L-N\epsilon}{M}-1$, and the property (ii) imply $\{Nx\}<\{Mx\}$. If $K=M$ happens, then we have $L=M$ and the contradiction $\{Nx\}=\{\frac{M-N\epsilon}{M}\}=1-\frac{N}{M}\epsilon>1-\epsilon=\{Mx\}$.

∎

Hereafter, we assume that integers $M,N$ satisfy $M>N>0$ and $(M,N)=1$ as in Theorem 8.1.

We write, for an integer $K$

[TABLE]

where $L$ is the integer defined by $L\equiv NK\bmod M$, and $N\leq L\leq M+N-1$. It is easy to see $\tau(0)=0,\tau(1)=1$ By Proposition 8.2, we have

[TABLE]

Proposition 8.3**.**

The mapping $\tau$ is the bijection from $\{1,2,\dots,M-1\}$ to $\Sigma:=\Sigma_{1}\cup\Sigma_{2}$, where $\Sigma_{1}:=\{\frac{1}{N},\dots,\frac{N-1}{N}\}$, $\Sigma_{2}:=\{\frac{1}{M-N},\dots,\frac{M-N}{M-N}\}$ and $\tau(K)=K$ holds in the ring $\mathbb{Z}/M\mathbb{Z}$. The set $\Sigma_{1}$ is empty if $N=1$. Moreover, we have

[TABLE]

and $K-\tau(K)$ $(K=1,\dots,M-1)$ are distinct.

Proof.

It is clear that the value $\tau(K)$ is in the set $\Sigma$ and $\#\Sigma=M-1$. Since $(M,M-N)=(M,N)=1$, we see that $\tau(K)=\frac{L}{N}=K$ in $\mathbb{Z}/M\mathbb{Z}$, which implies the injectivity of $\tau$. Write $NK=L+aM$ $(a\in\mathbb{Z})$. Suppose that $L>M$; it implies $K-\tau(K)=\frac{M}{N}(a+1)$, and from inequalities $N\leq L\leq M+N-1$ follows $\frac{N(K-1)}{M}-1+\frac{1}{M}\leq a\leq\frac{K-1}{M}N$, hence $-1<a<N$. If $a=N-1$ occurs, then $L=NK-(N-1)M$ holds and then the assumption $L>M$ implies $NK>NM$, which contradicts $K\leq M-1$. Next, suppose that $L<M$; then we see $K-\tau(K)=\frac{M}{M-N}(K-1-a)$ and inequalities $N\leq L\leq M+N-1$ imply $K-\frac{N}{M}(K-1)-1\leq K-1-a\leq(1-\frac{N}{M})K+\frac{N-1}{M}$, hence $0\leq K-1-a\leq(1-\frac{N}{M})(M-1)+\frac{N-1}{M}<M-N-1+2\frac{N}{M}-\frac{1}{M}<M-N+1$. If $K-1-a=M-N$ occurs, then the assumption $L<M$ implies $NK-aM<M$, hence $K-M+N=a+1>\frac{NK}{M}$, i.e. $(M-N)K>(M-N)M$, which is the contradiction. Next, assume that $K_{1}-\tau(K_{1})=K_{2}-\tau(K_{2})$ for $1\leq K_{1}<K_{2}\leq M-1$; it implies the contradiction $1\leq K_{2}-K_{1}=\tau(K_{2})-\tau(K_{1})<1$.

∎

When we change the domain of the mapping $\tau$ from $\{1,\dots,M-1\}$ to $\{0,1,\dots,M-1\}$, the set $\{0,1,\dots,M-1\}$ corresponds , through $K\to K-\tau(K)$ to $\Sigma^{\prime}:=\Sigma^{\prime}_{1}\sqcup\Sigma^{\prime}_{2}$ for

[TABLE]

where $K=0,1$ are supposed to correspond to $0\in\Sigma^{\prime}_{1},0\in\Sigma^{\prime}_{2}$, respectively. Now we see that

[TABLE]

Before the evaluation of the volume of the above, let us show the special case of $\tilde{n}=2$.

Proposition 8.4**.**

$I_{N,M}^{2}\cap\hat{\mathfrak{D}}_{2}$* is a finite set.*

Proof.

By (83), we have only to sow that $(\cup_{K=1}^{M-1}S_{\tau(K)})\cap\hat{\mathfrak{D}}_{2}$ is empty. Suppose that it is not empty and for $x_{i}=\frac{K_{i}-\tau(K_{i})+\epsilon_{i}}{M}$ with $0<\epsilon_{i}<\tau(K_{i}),1\leq K_{i}\leq M-1$ $(i=1,2)$, assume that $x_{1}+x_{2}\in\mathbb{Z}$. The equalities $0\leq\sum_{i=1}^{2}\frac{K_{i}-\tau(K_{i})}{M}<x_{1}+x_{2}<\sum_{i=1}^{2}\frac{K_{i}}{M}<2$ imply $x_{1}+x_{2}=1$, hence $K_{1}+K_{2}=M+\tau(K_{1})-\epsilon_{1}+\tau(K_{2})-\epsilon_{2}$. It implies $M<K_{1}+K_{2}<M+\tau(K_{1})+\tau(K_{2})\leq M+2$, that is $K_{1}+K_{2}=M+1$, hence

[TABLE]

Let $L_{i}$ $(i=1,2)$ be integers defining $\tau(K_{i})$, that is $L_{i}\equiv NK_{i}\bmod M,N\leq L_{i}\leq M+N-1$; then we have $L_{1}+L_{2}\equiv N(K_{1}+K_{2})\equiv N(M+1)\equiv N\bmod M$. Writing $L_{1}+L_{2}=N+kM$ $(k\in\mathbb{Z})$, we see $N\leq L_{2}=N+kM-L_{1}\leq M+N-1$ and $L_{1}\leq kM\leq L_{1}+M-1\leq 2M+N-2$, which implies $k=1,2$. First, suppose $k=1$; then $L_{1}+L_{2}=N+M$ and $0\leq L_{1}-N=M-L_{2}$, hence $\tau(K_{1})=\frac{M-L_{1}}{M-N}$ by definition of $\tau$, similarly $\tau(K_{2})=\frac{M-L_{2}}{M-N}$, which imply $\tau(K_{1})+\tau(K_{2})=1$ and so the contradiction $\epsilon_{1}+\epsilon_{2}=0$ by (87). Next, suppose $k=2$; we see that $L_{1}=(N+M-L_{2})+M>M$ and $\tau(K_{1})=\frac{L_{1}-M}{N}$, and similarly $\tau(K_{2})=\frac{L_{2}-M}{N}$. They also imply the contradiction $\tau(K_{1})+\tau(K_{2})=1$.

∎

Proposition 8.5**.**

Let $f(x)$ be a monic integral polynomial without rational roots, and let $n$ be a non-zero integer, and $A$ a positive number. If a prime $p$ is sufficiently large, then for any integer $r$ satisfying $f(r)\equiv 0\bmod p$, $\{\frac{nr}{p}\}\geq\frac{A}{p}$ holds.

Proof.

Suppose that for a prime $p$ and an integer $r$, $f(r)\equiv 0\bmod p$ and $\{\frac{nr}{p}\}<\frac{A}{p}$ hold. We may assume that $0\leq r<p$, and writing

[TABLE]

we see that $|k|p<(|n|+1)p$, and $0\leq R<A$ by the assumption $\left\{\frac{nr}{p}\right\}<\frac{A}{p}$, hence the possibility of $k,R$ is finite. Assume that there are infinitely many such primes $p$; then for some $k_{0},R_{0}$ there are infinitely many primes $p$ such that $nr=k_{0}p+R_{0}$ with $f(r)\equiv 0\bmod p$. Then, for the integral polynomial $g(x):=n^{\deg f}f(\frac{x}{n})$, we see that $g(R_{0})\equiv g(nr)\equiv n^{\deg f}f(r)\equiv 0\bmod p$, that is $g(R_{0})=0$, which is the contradiction. ∎

Proposition 8.6**.**

Let $f(x)=x^{2}-ax+b$ be an integral irreducible polynomial, and let $m,n$ be non-zero distinct integers. Then there are only finitely many primes $p$ such that $f(x)\equiv 0\bmod p$ has two roots $r_{1},r_{2}\in\mathbb{Z}$ such that $r_{1}\not\equiv r_{2}\bmod p$

[TABLE]

Proof.

Suppose that the inequalities (2) for a sufficiently large prime $p$; then by $r_{1}+r_{2}\equiv a\bmod p$ we have $\left\{\frac{nr_{2}}{p}\right\}<\left\{\frac{mr_{2}}{p}\right\}\Leftrightarrow$ \left\{\frac{n(a-r_{1})}{p}\right\}<\left\{\frac{m(a-r_{1})}{p}\right\}$$\Leftrightarrow\left\{\frac{na}{p}-\left\{\frac{nr_{1}}{p}\right\}\right\}<\left\{\frac{ma}{p}-\left\{\frac{mr_{1}}{p}\right\}\right\}\Leftrightarrow $\left\{\frac{nr_{1}}{p}\right\}-\frac{na}{p}>\left\{\frac{mr_{1}}{p}\right\}\linebreak[3]-\frac{ma}{p}$ by Proposition 8.5. Suppose that there are infinitely many primes satisfying (88). Then, writing $nr_{1}=kp+R,\,mr_{1}=k^{\prime}p+R^{\prime}$ $(k,k^{\prime}\in\mathbb{Z},0\leq R,R^{\prime}<p)$, the above inequality and the assumption (2) imply $R-na>R^{\prime}-ma$, i.e. $(n-m)a<R-R^{\prime}=p(\{\frac{nr_{1}}{p}\}-\{\frac{mr_{1}}{p}\})<0$. Therefore there is an integer $r_{0}$ with $(n-m)a<r_{0}<0$ such that $(n-m)r_{1}\equiv R-R^{\prime}\equiv r_{0}\bmod p$ for infinitely many primes $p$. Writing $g(x)=(n-m)^{\deg f}f(\frac{x}{n-m})\in\mathbb{Z}$, we have $g(r_{0})\equiv(n-m)^{\deg f}f(r_{1})\equiv 0\bmod p$ for infinitely many primes $p$, which contradicts the irreducibility of the polynomial $f(x)$.

∎

Let us begin the evaluation of the volume. Write $\tilde{M}(x):=\max(0,x)$ for simplicity.

Lemma 8.4**.**

Let $a,b,m$ be real numbers and suppose $a\leq b$ and $m\neq-1$. Then we have

[TABLE]

Proof.

By writing $t-w=W$, the left hand is equal to

[TABLE]

∎

Here we introduce the notation, for $x,\tau_{1},\dots,\tau_{n}\in\mathbb{R}$

[TABLE]

where $S$ runs over all $2^{n}$ subsets of $\{1,\dots,n\}$ and $|S|=\#S$. It is obvious that

[TABLE]

for any permutation $\sigma$. Moreover, we see

[TABLE]

In particular, the equation (90) says that

[TABLE]

Proposition 8.7**.**

For positive numbers $\tau_{1},\dots,\tau_{n}$, the volume of the set

[TABLE]

is

[TABLE]

Proof.

We use the induction on $n$. Write $U_{n}(x)=vol(V_{n}(x;\tau_{1},\dots,\tau_{n}))$ simply. For $n=1$, it is easy to see that

[TABLE]

Supposing (92), we see that $U_{n+1}(x)$ is

[TABLE]

∎

Example It is easy to check, by drawing the figure

[TABLE]

We note that the proposition implies the following equation :

[TABLE]

where $\tau_{1},\dots,\tau_{n}$ and $c$ are variables and $S$ runs over all $2^{n}$ subsets of $\{1,\dots,n\}$.

Proposition 8.8**.**

For positive numbers $\tau_{1},\dots,\tau_{n}$, the $n-1$-dimensional volume of the set

[TABLE]

is

[TABLE]

Proof.

Using orthnormal basis

[TABLE]

and the transformation $(y_{1},\dots,y_{n})=(x_{1},\dots,x_{n})({}^{t}\bm{f}_{1},\dots,{}^{t}\bm{f}_{n})$, that is $\sqrt{n}y_{n}=x_{1}+\dots+x_{n}$, we see, denoting $S_{n}(x;\tau_{1},\dots,\tau_{n})$ by $S_{n}(x)$ simply

[TABLE]

hence $U_{n}^{\prime}(x)=\sqrt{n}^{-1}vol(S_{n}(x))$, i.e. $vol(S_{n}(x))=\sqrt{n}\,U_{n}^{\prime}(x)$.

∎

Example The length of $S_{2}(x;\tau_{1},\tau_{2})$ is easily checked to be

[TABLE]

We note that the volume of $S_{n}(x;\tau_{1},\dots,\tau_{n})$ is positive if and only if $0<x<\sum\tau_{i}$ holds, and that $vol(S_{n}(x;\tau_{1},\dots,\tau_{n}))=vol(S_{n}(x;\tau_{\sigma(1)},\dots,\tau_{\sigma(n)}))$ for every permutation $\sigma\in S_{n}$, since a permutation induces the orthogonal transformation. In particular, for positive numbers $\tau_{1},\dots,\tau_{n}$ and a real number $x$, $W_{n-1}(x;\tau_{1},\dots,\tau_{n})>0$ if and only if $0<x<\sum\tau_{i}$ holds.

Now we see that, using (79), (86)

[TABLE]

where $\tau_{j}=\tau(K_{j})$, and note that

[TABLE]

which implies $0\leq l\leq\tilde{n}$, and continuing the above

[TABLE]

where $S$ runs over all subsets of $\{1,\dots,\tilde{n}\}$. Thus we see that

[TABLE]

where $K_{j}-\tau_{j}$ runs over $\Sigma^{\prime}:=\Sigma^{\prime}_{1}\sqcup\Sigma^{\prime}_{2}$ as stated after Proposition 8.3 and continuing

[TABLE]

by (80). For an integer $k$ $(0\leq k\leq\tilde{n})$, we see that

[TABLE]

where

[TABLE]

and

[TABLE]

Therefore we see that

[TABLE]

and so

[TABLE]

where

[TABLE]

We see that $\Delta_{0}(f)=f(0)=1,\Delta_{0}(g)=g(0)=0,\Delta_{1}(g)=\frac{1}{2}$, and

[TABLE]

thus

[TABLE]

Next, for an integer $k$ $(0\leq k\leq\tilde{n})$, let us evaluate

[TABLE]

where $\frac{\tau^{\prime}_{j}}{M}$ runs over the set $\Sigma^{\prime\prime}:=\Sigma^{\prime\prime}_{1}\sqcup\Sigma^{\prime\prime}_{2}$ for

[TABLE]

First of all, we note that, for positive integers $d,d_{1},d_{2}$ with $(d_{1},d_{2})=1$

[TABLE]

Because,

[TABLE]

and

[TABLE]

We see that

[TABLE]

Hence, we see that

[TABLE]

by (95) and for $1\leq k<\tilde{n}$ we have

[TABLE]

and

[TABLE]

Thus we have

[TABLE]

where we replaced $(-1)^{\tilde{n}}$ by $1$ by virtue of $B_{n}(0)=0$ for odd $n>1$. These complete the proof of Theorem 8.1.

9 Other direction

We know that for a prime $p$, the condition $p\in Spl(f)$ and the condition that $f(x)$ has $n(=\deg f(x))$ roots in the local field $\mathbb{Q}_{p}$ are equivalent except finitely many primes. So, it may be natural to ask, for a natural number $s$ how the distribution of roots of $f(x)\equiv 0\bmod p^{s}$ is, how the distribution of the coefficient of $p^{s}$ of the standard $p$-adic expansion of roots in $\mathbb{Q}_{p}$ is, and relations among them of several $s$’s.

With respect to the distribution of roots with a congruence condition, we may consider the distribution of roots $r_{i,s}$ for a positive integer $s$ such that

[TABLE]

with

[TABLE]

This problem is somewhat easy compared with the previous one, judging from the case of degree $1$.

We fixed the integer $s$ in the above, but contrary to it, we may ask the density of $s$ ($s\to\infty$) such that there exist integers $r_{i,s}$ satisfying above equations for a fixed prime $p\in Spl(f)$, $L,R_{i}$ (cf. 3.2 in [K6]). To make data in this case by computer, it takes much time.

One may ask about the equi-distribution similar to Conjecture 2 with congruence conditions.