Hasse-Minkowski theorem for quadratic forms on groups

TL;DR

This paper extends the Hasse-Minkowski theorem to quadratic forms on certain groups related to number fields, proving it holds for ranks 2 and 3 but not higher, thus generalizing classical results.

Contribution

It establishes the validity of the Hasse-Minkowski theorem for quadratic forms of rank 2 and 3 on specific algebraic groups, and provides counterexamples for higher ranks.

Findings

Hasse-Minkowski theorem holds for rank 2 and 3 forms.

Counterexamples exist for ranks higher than 3.

Generalizes classical theorem for binary and ternary forms.

Abstract

Consider groups such as Mordell-Weil groups of abelian varieties over number fields, odd algebraic $K$-theory groups of number fields, or finitely generated subgroups of the multiplicative groups of number fields. They are all equipped with systems of reduction maps; thus, one can investigate the Hasse-Minkowski theorem for quadratic forms with coefficients in such groups. In this paper, we prove that the theorem holds for the forms whose rank equals $2$ or $3$, and we demonstrate that it does not hold for higher ranks by providing a counterexample. We also show that our results constitute a generalization of the classic Hasse-Minkowski theorem for binary and ternary integral forms.

Equations

$$x^{2}P+y^{2}Q=T\mathrm{mod}v.$$

$$x^{2}P+y^{2}Q=T.$$

$$\begin{array}[]{ccc}l^{2}\parallel\mathrm{ord}_{v}P&\mathrm{and}&l^{3}\parallel\mathrm{ord}_{v}Q.\end{array}$$

$$\begin{array}[]{ccc}l^{2}\parallel\mathrm{ord}_{v}(x^{2}P)&\mathrm{and}&l^{3}\parallel\mathrm{ord}_{v}(y^{2}Q);\end{array}$$

$$\begin{array}[]{ccc}l\nmid\mathrm{ord}_{v}(x^{2}P)&\mathrm{and}&l^{3}\parallel\mathrm{ord}_{v}(y^{2}Q);\end{array}$$

$$\begin{array}[]{ccc}l^{2}\parallel\mathrm{ord}_{v}(x^{2}P)&\mathrm{and}&l^{2}\nmid\mathrm{ord}_{v}(y^{2}Q);\end{array}$$

$$(b{x}^2-a{y}^2)P=bT\mathrm{mod}v.$$

$$x^{2}P+y^{2}Q+z^{2}R=T\mathrm{mod}v.$$

$$x^{2}P+y^{2}Q+z^{2}R=T.$$

$$\begin{array}[]{ccccc}2^{1+e}\parallel\mathrm{ord}_{v}(P+Q)&,&2^{3+e}\parallel\mathrm{ord}_{v}Q&,&2^{4+e}\parallel\mathrm{ord}_{v}(Q+R).\end{array}$$

$$\begin{array}[]{ccccc}2^{3+e}\parallel\mathrm{ord}_{v}P&,&2^{3+e}\parallel\mathrm{ord}_{v}Q&,&2^{4+e}\parallel\mathrm{ord}_{v}R.\end{array}$$

$$\begin{array}[]{cccc}\mathrm{either}&2^{1+e}\nmid\mathrm{ord}_{v}(z^{2}R)&\mathrm{or}&2^{2+e}\parallel\mathrm{ord}_{v}(z^{2}R)\end{array}$$

$$\begin{array}[]{cccc}\mathrm{either}&2^{1+e}\nmid\mathrm{ord}_{v}(x^{2}P+y^{2}Q)&\mathrm{or}&2^{2+e}\parallel\mathrm{ord}_{v}(x^{2}P+y^{2}Q).\end{array}$$

$$\begin{array}[]{ccc}2^{1+e}\parallel\mathrm{ord}_{v}(y^{2}(P+Q))&\mathrm{and}&2^{1+e}\nmid\mathrm{ord}_{v}((x^{2}-y^{2})P).\end{array}$$

$$\begin{array}[]{ccc}2^{3+e}\parallel\mathrm{ord}_{v}(x^{2}P)&\mathrm{and}&2^{3+e}\nmid\mathrm{ord}_{v}(y^{2}Q)\end{array}$$

$$(t{x}^2+s{y}^2)Q+s{z}^{2}R=sT\mathrm{mod}v.$$

$$(wt{x}^2+ws{y}^2+us{z}^2)Q=wsT\mathrm{mod}v.$$

$$aP+bQ+cR=0$$

$$(a{y}^2-b{x}^2)Q+(a{z}^2-c{x}^2)R=aT\mathrm{mod}v.$$

$$\begin{array}[]{ccc}l^{2}\parallel\mathrm{ord}_{v}Q&\mathrm{and}&l\parallel\mathrm{ord}_{v}R.\end{array}$$

$$l\mid (a{y}^2-b{x}^2)$$

$$a(sw{)}^{2}P+a(tw{)}^{2}Q+a(su{)}^{2}R=0.$$

$$(2x_1^2+x_2^2+x_3^2+x_4^2+\dots +x_n^2)P.$$

$$(2x_1^2+x_2^2+x_3^2+x_4^2+\dots +x_n^2)P=0\mathrm{mod}v.$$

Full text

1. Introduction.

Let $V$ denote all places on $\mathbb{Q}$ including $\infty$. The Hasse-Minkowski theorem (cf. [H1]) states that a quadratic form over $\mathbb{Q}$ represents [math] if and only if it represents [math] in $\mathbb{Q}_{v}$ for every $v\in V$. Since quadratic forms can be defined on any module over a commutative ring it is a natural question whether we can obtain an analogue of Hasse-Minkowski theorem provided our module is equipped with a system of reduction maps.

A perfect example of such module is the Mordell-Weil group of an elliptic curve over a number field. Observe that here not every place defines a reduction map since we have to exclude primes of bad reduction. Note however that the original Hasse-Minkowski theorem also has its stronger versions where some number of places can be omitted (see Proposition 4); our proofs depend on those results.

The groups for which we prove Hasse-Minkowski theorem in this paper are the following Mordell-Weil type groups:

1. (1)

${R}_{F,S}^{\times}$, $S$-units groups, where $F$ is a number field and $S$ is a finite set of ideals in the ring of integers ${R}_{F}$, 2. (2)

$A(F)$, Mordell-Weil groups of abelian varieties over number fields $F$ with $\mathrm{End}_{\bar{F}}(A)=\mathbb{Z}$, 3. (3)

$K_{2n+1}(F)$, $n>0$, odd algebraic $K$-theory groups.

To deal with them it is enough for us to introduce the following abstract nonsense axiomatic setup:

Let $B$ be an abelian group and $r_{v}\colon B\to B_{v}$ be an infinite family of groups homomorphisms whose targets $B_{v}$ are finite abelian groups. We will use the following notation:

$P\mod v$ denotes $r_{v}(P)$ for $P\in B$

$P=Q\mod v$ means $r_{v}(P)=r_{v}(Q)$ for $P,Q\in B$

$\Lambda_{\mathrm{tors}}$ the torsion part of a subgroup $\Lambda<B$

$\mathrm{ord}\ T$ the order of a torsion point $T\in B$

$\mathrm{ord}_{v}P$ the order of a point $P\mod v$

$l^{k}\parallel n$ means that $l^{k}$ exactly divides $n$, i.e. $l^{k}\mid n$ and $l^{k+1}\nmid n$

where $l$ is a prime number, $k$ a positive integer

and $n$ a natural number.

We impose the following two assumptions on the family $r_{v}\colon B\to B_{v}$:

Assumptions.

1. (1)

Let $l$ be a prime number and $(k_{1},\ldots,k_{m})$ a sequence of nonnegative integers. If $P_{1},\ldots,P_{m}\in B$ are points linearly independent over ${\mathbb{Z}}$ then there is a positive density set of primes $v$ in $F$ such that $l^{k_{i}}\parallel\mathrm{ord}_{v}P_{i}$ if $k_{i}>0$ and $l\nmid\mathrm{ord}_{v}P_{i}$ if $k_{i}=0$. 2. (2)

For almost all $v$ the map $B_{\mathrm{tors}}\to B_{v}$ is injective.

(For the families of groups we described above the axioms are valid, in particular Assumption 1 is fulfilled by [Bar], Theorem 5.1 and Assumption 2 by [BGK], Lemma 3.11. The validity of the technical hypothesis 111note the misprint in its formulation: $\rho(G_{F})$ should read $\rho_{l}(G_{F})$ from the first sentence of Theorem 5.1 in [Bar] is clear in the $K$-theory groups case (resp. ${R}_{F,S}^{\times}$ case) since here $\rho_{l}$ is simply one-dimensional representation given by the $(n+1)$th tensor power of cyclotomic character (resp. by cyclotomic character); in abelian varieties case it is asserted by Corollary 1 in [Bog].)

2. Quadrartic forms of rank $2$.

Theorem 1**.**

Let $P,Q\in B$ be points of infinite order. The following are equivalent:

• •

For almost every $v$ there exist coprime integers $x,y$ and a point $T\in B_{\mathrm{tors}}$ such that

[TABLE]

• •

There exist coprime integers $x,y$ and a point $T\in B_{\mathrm{tors}}$ such that

[TABLE]

Proof.

Suppose that $P,Q$ are linearly independent. Fix a prime number $l$ coprime to $\#B_{\mathrm{tors}}$. By Assumption 1 there is a positive density set of primes $v$ such that

[TABLE]

Thus if both $x,y$ are coprime to $l$ then

[TABLE]

if $l\mid x$ and $l\nmid y$ then

[TABLE]

and if $l\nmid x$ and $l\mid y$ then

[TABLE]

in all cases we get a contradiction to (1). Hence $P,Q$ are linearly dependent.

Write $aP+bQ=0$ for nonzero integers $a,b$. Multiplying (1) by $b$ we get

[TABLE]

Fix an arbitrary prime number $l$ coprime to $\#B_{\mathrm{tors}}$. By Assumption 1 there is a positive density set of primes $v$ such that $l\parallel\mathrm{ord}_{v}P$. So by (3) there are coprime integers $x,y$ such that $l\mid(bx^{2}-ay^{2})$. Thus by Proposition 4 (a) the form $bx^{2}-ay^{2}$ represents [math]. Putting $bx^{2}-ay^{2}=0$ we get $b(x^{2}P+y^{2}Q)=0.$

∎

3. Quadrartic forms of rank $3$.

Theorem 2**.**

Let $P,Q,R\in B$ be points of infinite order. The following are equivalent:

• •

For almost every $v$ there exist integers $x,y,z$ with $\gcd(x,y,z)=1$ and a point $T\in B_{\mathrm{tors}}$ such that

[TABLE]

• •

There exist integers $x,y,z$ with $\gcd(x,y,z)=1$ and a point $T\in B_{\mathrm{tors}}$ such that

[TABLE]

Proof of Theorem 2.

First we prove that $P,Q,R$ are linearly dependent. Indeed, assume that $P,Q,R$ are linearly independent. This means that the points $P+Q,Q,Q+R$ are also independent. Let $2^{e}$ be the maximal power of $2$ dividing $\#B_{\mathrm{tors}}$. By Assumption 1 there is a positive density set of primes $v$ such that

[TABLE]

That gives

[TABLE]

Now by (4) we have $2\mid z$ thus

[TABLE]

so respectively

[TABLE]

Since $\gcd(x,y,z)=1$ we have to consider two cases: when both $x,y$ are odd and when exactly one of $x,y$ is even.

If both $x,y$ are odd then $8\mid(x^{2}-y^{2})$ so we get by (6) and (7) that

[TABLE]

Since $x^{2}P+y^{2}Q=y^{2}(P+Q)+(x^{2}-y^{2})P$ we have $2^{1+e}\parallel\mathrm{ord}_{v}(x^{2}P+y^{2}Q)$ but that contradicts (9).

If $x$ is odd and $y$ is even then

[TABLE]

but that together with (8) contradicts (4). If $y$ is odd and $x$ is even then we argue in the same manner.

We have proven that $P,Q,R$ are linearly dependent.

First we analyse the case when among the pairs $\left\{P,Q\right\}$, $\left\{P,R\right\}$, $\left\{Q,R\right\}$ one is a pair of linearly dependent points. Without loss of generality we can assume that this pair is $\left\{P,Q\right\}$, i.e., there are nonzero rational integers $s,t$ such that $sP=tQ$. If $-s/t$ is the square of a rational number then we are done. Indeed, write $-s/t=(p/q)^{2}$ with coprime $p,q$. Then $t(p^{2}P+q^{2}Q)=0$ so we can put $(x,y,z)=(p,q,0)$ in (5).

So suppose that $-s/t$ is not a square. Multiplying (4) by $s$ we get

[TABLE]

Fix a prime number $l$ such that $l$ is coprime to both $s,t$ and to $\#B_{\mathrm{tors}}$ and such that $-s/t$ is a quadratic nonresidue modulo $l$. Suppose that $Q,R$ are linearly independent. By Assumption 1 there is a positive density set of primes $v$ such that $l^{2}\parallel\mathrm{ord}_{v}Q$ and $l^{3}\parallel\mathrm{ord}_{v}R$. Then (10) implies that $l\mid z$. This means that $l^{2}\nmid\mathrm{ord}_{v}(sz^{2}R)$ so again by (10) we get $l\mid(tx^{2}+sy^{2})$. But $\gcd(x,y,z)=1$ hence both $x,y$ are coprime to $l$ and we can write $(x/y)^{2}=-s/t\mod l$ contrary to our assumption that $-s/t$ is nonresidue. Thus $Q,R$ are linearly dependent.

Write $uQ=wR$ for some nonzero rational integers $u,w$. Multiplying (10) by $w$ we get

[TABLE]

Denote by $\mathcal{P}$ the set of all prime divisors of $\#B_{\mathrm{tors}}$. For every $p\in\mathcal{P}$ let $p^{e_{p}}$ be the maximal power of $p$ dividing $\#B_{\mathrm{tors}}$. Define $e=\max\left\{e_{p}:p\in\mathcal{P}\right\}$. Fix an arbitrary prime number $l$ and a positive integer $k$. By Assumption 1 there is a positive density set of primes $v$ such that $l^{k+e}\parallel\mathrm{ord}_{v}Q$ so $l^{k}\mid(wtx^{2}+wsy^{2}+usz^{2})$ by (11). Thus the quadratic form $(wtx^{2}+wsy^{2}+usz^{2})$ represents 0 modulo arbitrary $l^{k}$ and hence by Proposition 4 (b) it represents 0.

Now we turn to the case when among the pairs $\left\{P,Q\right\}$, $\left\{P,R\right\}$, $\left\{Q,R\right\}$ none is a pair of linearly dependent points.

Write

[TABLE]

with $abc\neq 0$. Multiplying (4) by $a$ and (12) by $x^{2}$ and subtracting the results we get

[TABLE]

Fix a prime number $l$ coprime to the numbers $a,b,c,\#B_{\mathrm{tors}}$. By Assumption 1 there is a positive density set of primes $v$ such that

[TABLE]

By (13) we get

[TABLE]

thus if $l$ divides either $x$ or $y$ then it divides them both. But that implies by (14) and (13) that $l\nmid\mathrm{ord}_{v}(az^{2}R)$ so we have $l\mid z$ which is a contradiction to our assumption that $\gcd(x,y,z)=1$. Thus $l$ divides neither $x$ nor $y$ and (15) means that $b/a$ is a square modulo $l$.

The only restriction on $l$ in the above argument is that it cannot divide the numbers $a,b,c,\#B_{\mathrm{tors}}$ hence it is valid for all but finitely many prime numbers $l$ and we can use Proposition 4 (a) which asserts that $b/a$ is the square of a rational number. Analogously we prove the same for $c/a$.

Hence we can put $b/a=(t/s)^{2}$ and $c/a=(u/w)^{2}$ and by (12) we get

[TABLE]

∎

4. Quadratic forms of higher ranks. Auxiliary results.

Theorems 1 and 2 cannot be extended directly for quadratic forms of higher ranks, even if we assume that they represent zero for all primes of good reduction. Indeed, we have the following result:

Proposition 3**.**

Fix $n\geq 4$. Let $P$ be a point of infinite order. Define the form $f$ by

[TABLE]

Then $f=0$ if and only if $(x_{1},\ldots,x_{n})=(0,\ldots,0)$ but for every prime $v$ of good reduction there exist integers $x_{1},\ldots,x_{n}$ with $\gcd(x_{1},\ldots,x_{n})=1$ such that

[TABLE]

Proof.

The form $(2x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+\ldots+x_{n}^{2})$ is positive and the point $P$ is of infinite order so $(2x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+\ldots+x_{n}^{2})P=0$ if and only if $(x_{1},\ldots,x_{n})=(0,\ldots,0)$.

Let $k$ be the order of $P\mod v$. It is an easy corollary of the theorem of Gauss on sums of three squares (cf. [Sie], page 395, exercise 4) that $2k$ is of the form $2a^{2}+b^{2}+c^{2}+1$ where $a,b,c$ are integers. Thus in order to get (16) we can put $x_{1}=a,x_{2}=b,x_{3}=c,x_{4}=1$ and $x_{i}=0$ for $i>4$. ∎

Proposition 4**.**

1. (a)

Let $f$ be an integral quadratic form of rank $2$. If $f\mod v$ represents [math] for all but finitely many $v\in V$ then $f$ represents [math]. 2. (b)

Let $f$ be an integral quadratic form of rank $3$. If $f$ represents [math] in $\mathbb{Q}_{v}$ for all except at most one $v\in V$ then $f$ represents [math].

Proof.

(a) This statement is an immediate corollary of the theorem proved in [H2].

(b) Cf. [S], p. 43, Corollary 3. ∎