Roots of unity and unreasonable differentiation

TL;DR

This paper investigates the conditions under which differentiation of polynomials at roots of unity is valid within modular arithmetic frameworks, revealing insights into algebraic structures and computational limitations.

Contribution

It introduces criteria for legal differentiation at roots of unity in modular settings, advancing understanding of polynomial behavior in finite fields.

Findings

Derived conditions for differentiation validity at roots of unity

Identified limitations of modular differentiation methods

Provided theoretical foundations for algebraic computations

Abstract

We explore when it is legal to differentiate a polynomial evaluated at a root of unity using modular arithmetic.

Equations

$$f^{(k)}(\alpha )=\sum _{i=0}^{n-1}{i}^{\underline{k}}{\alpha }^{i-k}\equiv 0(\mathrm{mod}n).$$

$$\sum _{i=n_0}^{n_1-1}{i}^{\underline{k}}=\sum _{i=n_0}^{n_1-1}\Delta \left(\frac{i^{\underline{k+1}}}{k+1}\right)=\frac{n_1^{\underline{k+1}}}{k+1}-\frac{n_0^{\underline{k+1}}}{k+1}=\frac{n_1^{\underline{k+1}}-n_0^{\underline{k+1}}}{k+1}.$$

$$\sum _{i=0}^{n-1}{i}^{\underline{k}}{\alpha }^{i-k}\equiv \frac{(n-1{)}^{\underline{k}}}{k+1}n(\mathrm{mod}{p}^{\ell })\text{where ℓ:=νp(n).}$$

$$\sum _{i=0}^{n-1}{i}^{\underline{k}}{\alpha }^{i-k}$$

$$=\sum _{j=0}^{n-1-k}{y}^j\sum _{i=k+j}^{n-1}{i}^{\underline{k}}\left(\genfrac{}{}{0ex}{}{i-k}{j}\right)=\sum _{j=0}^{n-1-k}\frac{y^j}{j!}\sum _{i=k+j}^{n-1}{i}^{\underline{k+j}}$$

$$=\sum _{j=0}^{n-1-k}\frac{y^j}{j!}\left(\frac{n^{\underline{k+j+1}}}{k+j+1}-\frac{(k+j{)}^{\underline{k+j+1}}}{k+j+1}\right)$$

$$=\sum _{j=0}^{n-1-k}\frac{y^j}{j!}\frac{n^{\underline{k+j+1}}}{k+j+1}=\sum _{j=0}^{n-1-k}\frac{y^j}{j!}\frac{(n-1{)}^{\underline{k+j}}}{k+j+1}n.$$

$$\sum _{i=0}^{n-1}{i}^{\underline{k}}{\alpha }^{i-k}\equiv \frac{(n-1{)}^{\underline{k}}}{k+1}n(\mathrm{mod}{p}^{\ell }).\text{qed}$$

$$f^{(k)}(t)=\sum _{i=0}^{n-1}{i}^{\underline{k}}{t}^{i-k}=f_1^{(0)}(t)f_2^{(k)}(t)-\sum _{i=0}^{k-1}\left(\genfrac{}{}{0ex}{}{k}{i}\right)n^{\underline{k-i}}{t}^{n-k+i}i!(1-t{)}^{-1-i}.$$

$$\sum _{i=0}^{n-1}{i}^{\underline{k}}{t}^{i-k}=-(t^n-1)k!(1-t{)}^{-1-k}-\sum _{i=0}^{k-1}{k}^{\underline{i}}{n}^{\underline{k-i}}{t}^{n-k+i}(1-t{)}^{-1-i}.$$

Full text

Acknowledgements: SG gratefully acknowledges support from the Australian Research Council Discovery Project DP190100450.

2010 Math Subject Classification: 11A07, 11-01.

1. Sometimes legal operations

The equation $(x+y)^{p}=x^{p}+y^{p}$ is valid in a field of prime characteristic $p$. Thus an apparent error can be a legitimate deduction in the right circumstances.

Denote the $k$th derivative of $t^{n}$ as $n^{\underline{k}}t^{n-k}$ where $n^{\underline{k}}$ equals $n(n-1)\cdots(n-k+1)$ for $k>0$, and 1 if $k=0$. We pronounce $n^{\underline{k}}$ as “$n$ falling factorial $k$”. Observe that $n^{\underline{k}}$ is divisible by each integer in $\{1,\dots,k\}$, and $n^{\underline{k}}=0$ for $n<k$. Also the $k$th derivative $f^{(k)}(t)$ of a power series $f(t)=\sum_{n\geqslant 0}f_{n}t^{n}$ equals $\sum_{n\geqslant 0}f_{n}n^{\underline{k}}t^{n-k}=\sum_{n\geqslant k}f_{n}n^{\underline{k}}t^{n-k}$.

Let $\alpha\in\mathbb{Z}$ satisfy $\alpha^{n}\equiv 1\pmod{n}$. Thus $\alpha$ is a root, modulo $n$, of the polynomial $t^{n}-1=(t-1)f(t)$, where $f(t)=\sum_{i=0}^{n-1}t^{i}$. It is clear that $f^{(0)}(\alpha)\equiv 0\pmod{n}$ when $\alpha\equiv 1\pmod{n}$. However, it seems unreasonable to expect that $f^{(k)}(\alpha)=\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}\equiv 0\pmod{n}$ holds for all $k\geqslant 0$. What looks like a blunder turns out to be true under the (unreasonably) weak assumptions of Theorem 1.

Theorem 1**.**

Suppose $k\geqslant 0$, $n\geqslant 1$, $\alpha\in\mathbb{Z}$ where $\alpha^{n}\equiv 1\pmod{n}$. Then

[TABLE]

if and only if at least one of the following hold:

1. (a)

$k+1\not\in\{4,q\}$* where $q$ is prime, or* 2. (b)

$k+1=4$* and $4\nmid n$, or* 3. (c)

$k+1$* is a prime $q$, and $q\nmid n$ or $\alpha\not\equiv 1\pmod{q}$.*

The motivation for Theorem 1 came from a (presently unfinished [BDG]) study of input-output automata on a group $G$. We considered the finite groups $G$ for which there exists a ‘constant’ $k\in G$ and a function $f\colon G\to G$ satisfying $f(xk)=xf(x)$ for all $x\in G$. We call these $J$-groups (as they are related to the Jacobson radical of a near ring). A simple argument shows that $J$-groups must have odd order, and hence are solvable by [FT]. We conjectured [BDG] that any nilpotent group of odd order is a $J$-group. To prove that many metacyclic groups are $J$-groups required the $k=0$ and $k=1$ cases of Theorem 1. The proof for all $k\geqslant 0$ is not much harder.

2. The proofs

We first establish some preliminary results before proving Theorem 1.

Henceforth, $n,i,j,k$ will be integers.

A sum $\sum_{i=n_{0}}^{n_{1}-1}g(i)$ collapses if we find a function $G$ such that $g(i)=G(i+1)-G(i)$ for $n_{0}\leqslant i<n_{1}$. Then $\sum_{i=n_{0}}^{n_{1}-1}g(i)=G(n_{1})-G(n_{0})$. By analogy with differentiation, we write $(\Delta G)(i)=G(i+1)-G(i)$. For example, if $g(i)=i^{\underline{k}}$, then it follows from $\Delta(i^{\underline{k+1}})=(i+1)i^{\underline{k}}-i^{\underline{k}}(i-k)=(k+1)i^{\underline{k}}$ that $G(i)=i^{\underline{k+1}}/(k+1)$. Hence

[TABLE]

Clearly $k$ divides $n^{\underline{k}}$ for all $n\geqslant 0$ and $k\geqslant 1$. The $p$-adic valuation $\nu_{p}(n)$ of an integer $n\neq 0$ is defined by $\nu_{p}(n)=\log_{p}(n_{p})$ where $n_{p}$ is the largest $p$-power divisor of $n$. This (additive) valuation extends to $\mathbb{Q}^{\times}$ by defining $\nu_{p}(r/s)$ to be $\nu_{p}(r)-\nu_{p}(s)$.

Lemma 2**.**

Suppose $k\geqslant 1$ and $n\geqslant 1$. Let $p\mid(k+1)$ where $p$ is a prime, and let $e=\nu_{p}(k+1)\geqslant 1$.

1. (a)

If $k+1\neq p^{e}$, then $\nu_{p}((n-1)^{\underline{k}})\geqslant e$. 2. (b)

If $k+1=p^{e}$, then $\nu_{p}((n-1)^{\underline{k}})\geqslant e-1$ where equality holds only if $p\mid n$. 3. (c)

$\nu_{p}((n-1)^{\underline{k}}/(k+1))<0$* if and only if $k+1\in\{4,p\}$ and $(k+1)\mid n$, in which case $\nu_{p}((n-1)^{\underline{k}}/(k+1))=-1$. *

Proof.

(a) Suppose first that $k+1$ is not a $p$-power and write $k+1=ab$ where $\gcd(a,b)=1$ and $1<a<b<k+1$. Since $a,b\leqslant k$ it follows that $a$ and $b$, and hence $k+1=ab$, divide $(n-1)^{\underline{k}}$. Hence $e=\nu_{p}(k+1)\leqslant\nu_{p}((n-1)^{\underline{k}})$. This proves (a).

(b) Suppose now that $k+1=p^{e}$. As $p^{e-1}\leqslant k$, we deduce that $p^{e-1}\mid(n-1)^{\underline{k}}$, and so $\nu_{p}((n-1)^{\underline{k}})\geqslant e-1$. Suppose $\nu_{p}((n-1)^{\underline{k}})=e-1$. As $k+1$ divides $n^{\underline{k+1}}=n(n-1)^{\underline{k}}$ but not $(n-1)^{\underline{k}}$, we deduce that $p$ divides $n$. This proves (b).

(c) Assume first that $\nu_{p}((n-1)^{\underline{k}}/(k+1))<0$, that is, $\nu_{p}((n-1)^{\underline{k}})<\nu_{p}(k+1)=e$. Part (a) implies $k+1=p^{e}$ and Part (b) implies $\nu_{p}((n-1)^{\underline{k}})=e-1$ and $p\mid n$, so that $\nu_{p}((n-1)^{\underline{k}}/(k+1))=-1$. Thus each factor of $(n-1)^{\underline{k}}$ of the form $n-jp$ with $1\leqslant j\leqslant p^{e-1}-1$ is a multiple of $p$, and so $\nu_{p}((n-1)^{\underline{k}})\geqslant p^{e-1}-1$. Therefore $p^{e-1}-1\leqslant e-1$, that is $p^{e-1}\leqslant e$. The latter inequality is true for $e=1$ and all primes $p$, and for $e=2$ and $p=2$, and false otherwise.

If $e=1$, then $k+1=p\mid n$. If $e=2$ and $p=2$, then $k+1=4$, $2\mid n$ and $\nu_{2}((n-1)^{\underline{3}})=1$. Thus $n-1$ and $n-3$ are odd while $n-2\equiv 2\pmod{4}$. It follows that $n\equiv 0\pmod{4}$, and so in both cases $k+1$ divides $n$.

Conversely, assume that $k+1\in\{4,p\}$ and $(k+1)\mid n$. If $k+1=4\mid n$, then $(n-1)^{\underline{k}}=(n-1)(n-2)(n-3)$ where $n-i\equiv 4-i\pmod{4}$. Thus $\nu_{2}((n-1)^{\underline{3}})=1<\nu_{2}(k+1)=2$. If $k+1=p\mid n$, then $(n-1)^{\underline{k}}=(n-1)(n-2)\cdots(n-p+1)$ where $n-k\equiv p-k\not\equiv 0\pmod{p}$. Thus, in both cases, we have $\nu_{p}((n-1)^{\underline{k}})=e-1<\nu_{p}(k+1)=e$, as desired. ∎

Corollary 3**.**

By Lemma 2, we have that $(n-1)^{\underline{k}}/(k+1)$ is an integer unless

1. (a)

$k+1=4$* and $4\mid n$, or* 2. (b)

$k+1$* is a prime $p$, and $p\mid n$.*

Moreover, $\nu_{p}((n-1)^{\underline{k}}/(k+1))\geqslant-1$ and $\nu_{p}((n-1)^{\underline{k}}/(k+1))\geqslant 0$ if $(k+1)\nmid n$.

To prove Theorem 1, we will also need the following lemma.

Lemma 4**.**

Suppose $k\geqslant 0$, $n\geqslant 1$ and $\alpha\equiv 1\pmod{p}$ where $p$ is prime. Then

[TABLE]

Proof.

Since $\alpha\equiv 1\pmod{p}$, we have $\alpha=1+y$ where $p\mid y$. Using $i^{\underline{k}}=0$ if $0\leqslant i<k$, and Eq. (2) gives:

[TABLE]

Consider the summands in (2) with $j\geqslant 1$. By Legendre’s formula, $\nu_{p}(j!)=\sum_{i=1}^{\infty}\lfloor\frac{j}{p^{i}}\rfloor$. (The sum is finite as $\lfloor\frac{j}{p^{i}}\rfloor=0$ for $p^{i}>j$.) Thus $\nu_{p}(j!)<\sum_{i=1}^{\infty}\frac{j}{p^{i}}=\frac{j}{p-1}$. Therefore $\nu_{p}(j!)<j$ for $j\geqslant 1$, and hence $p$ divides $y^{j}/j!$. Since $\nu_{p}(y^{j}/j!)\geqslant 1$ for all $j\geqslant 1$, and $\nu_{p}(\frac{(n-1)^{\underline{k+j}}}{k+j+1})\geqslant-1$ by Corollary 3, we have $\nu_{p}(\frac{y^{j}}{j!}\frac{(n-1)^{\underline{k+j}}}{k+j+1})\geqslant 0$. However $p^{\ell}\mid n$, and so $\frac{y^{j}}{j!}\frac{(n-1)^{\underline{k+j}}}{k+j+1}n\equiv 0\pmod{p^{\ell}}$ for each $j\geqslant 1$. Hence

[TABLE]

Proof of Theorem 1.

When $n=1$, Eq. (1) is trivially true. Moreover, one of (a), (b) or (c) is true when $n=1$ since $4\nmid n$ and $q\nmid n$ for any prime $q$.

We now assume $n>1$. Suppose that $n=n_{1}\cdots n_{r}$ where the $n_{j}$ are pairwise coprime and $n_{j}>1$ for each $j$. Given the ring isomorphism $\mathbb{Z}_{n}\to\mathbb{Z}_{n_{1}}\times\cdots\times\mathbb{Z}_{n_{r}}$, Eq. (1) holds if and only if $n_{j}\mid\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}$ for each $j$. Suppose that $n_{j}=p_{j}^{\ell_{j}}$ where each $p_{j}$ is prime. Fix a prime factor $p$ of $n$, and set $\ell:=\nu_{p}(n)$.

For each prime factor $p$ of $n$, we divide the proof in two cases.

Claim 1: If $\alpha\not\equiv 1\pmod{p}$, then $\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}\equiv 0\pmod{p^{\ell}}$.

Suppose that $\alpha\not\equiv 1\pmod{p}$. Consider the identity $f(t)=\sum_{i=0}^{n-1}t^{i}=f_{1}(t)f_{2}(t)$ where $f_{1}(t)=t^{n}-1$ and $f_{2}(t)=(t-1)^{-1}$. The $k$-fold derivative of the product $f_{1}f_{2}$ is $(f_{1}f_{2})^{(k)}=\sum_{i=0}^{k}\binom{k}{i}f_{1}^{(k-i)}f_{2}^{(i)}$ by Leibnitz’ formula. We have $f_{1}^{(i)}(t)=n^{\underline{i}}t^{n-i}$ for $i>0$, and $f_{2}^{(i)}(t)=(-1)^{i}i!(t-1)^{-1-i}=-i!(1-t)^{-1-i}$ for $i\geqslant 0$. Hence, for $t\neq 1$,

[TABLE]

Replacing $\binom{k}{i}i!$ with $k^{\underline{i}}$ gives

[TABLE]

Substituting $t=\alpha$ and noting that $\alpha^{n}\equiv 1\pmod{p^{\ell}}$ and $1-\alpha$ is a unit modulo $p^{\ell}$ shows that $\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}\equiv 0\pmod{p^{\ell}}$. The sum vanishes modulo $p^{\ell}$ for $k=0$, and for $k>0$ because $n$ divides $n^{\underline{k-i}}$ for $i<k$ and $p^{\ell}\mid n$. This proves Claim 1.

Claim 2: If $\alpha\equiv 1\pmod{p}$ and at least one of the conditions (a), (b), or (c) hold, then $\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}\equiv 0\pmod{p^{\ell}}$.

Suppose that $\alpha\equiv 1\pmod{p}$ and at least one of the conditions (a), (b), or (c) hold. We argue that $\frac{(n-1)^{\underline{k}}}{k+1}$ is an integer. Suppose not. Then by Corollary 3, condition (i) or (ii) holds. Since $\alpha\equiv 1\pmod{p}$ both  (i) and (ii) are incompatible with (a), (b), and (c). This shows that $\frac{(n-1)^{\underline{k}}}{k+1}$ is an integer, and hence Claim 2 is true by Lemma 4.

In summary, we have $n_{j}\mid\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}$ for each $j$, and so (1) holds if at least one of the conditions (a), (b), or (c) hold.

To finish the proof, we now prove the converse. Assume (1) holds but (a) does not hold. Then $n_{j}\mid\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}$ for each $j$, and $k+1\in\{4,q\}$ where $q$ is a prime. We must show that conditions (b) or (c) hold.

Suppose $k+1=4$. Assume $4\mid n$. Consider the prime factor $2$ of $n$. Since $\alpha^{n}\equiv 1\pmod{n}$, we have $\alpha\equiv 1\pmod{2}$. Thus Lemma 4 implies $\sum_{i=0}^{n-1}i^{\underline{3}}\alpha^{i-3}\equiv\frac{(n-1)^{\underline{3}}}{4}n\pmod{2^{\ell}}$ where $\nu_{2}(n)=\ell$. By Lemma 2(c), $\nu_{2}((n-1)^{\underline{3}}/4)<0$, and so $2^{\ell}$ does not divide $\frac{(n-1)^{\underline{3}}}{4}n$, a contradiction. Hence, if $k+1=4$, then $4\nmid n$ and (b) holds.

Suppose $k+1$ is a prime $q$ and $\alpha\equiv 1\pmod{q}$. Assume $q\mid n$. Consider the prime factor $q$ of $n$. Then $\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}\equiv\frac{(n-1)^{\underline{k}}}{k+1}n\pmod{q^{\ell}}$ where $\nu_{q}(n)=\ell$, by Lemma 4. However, Lemma 2(c) shows $\nu_{q}((n-1)^{\underline{k}}/(k+1))<0$, and so $q^{\ell}$ does not divide $\frac{(n-1)^{\underline{k}}}{k+1}n$, a contradiction. Hence, if $k+1=q$ with $\alpha\equiv 1\pmod{q}$, then $q\nmid n$ and condition (c) holds. ∎

Finally, observe that the requirement $\alpha\not\equiv 1\pmod{q}$ in Theorem 1(c) is needed. For example take $\alpha=5$, $n=6$, $k=2$. Then $\sum_{i=0}^{n-1}i^{\underline{k}}\alpha^{i-k}\equiv 0\pmod{n}$ and $q=k+1=3$. Thus conditions (a) and (b) do not hold, and the only part of (c) that holds is $\alpha\not\equiv 1\pmod{q}$. In condition (b), we do not need to add “or $\alpha\not\equiv 1\pmod{2}$” because that never happens when $4\mid n$.

Acknowledgement. We thank the referee for their helpful suggestions.