Hasse-Minkowski theorem for quadratic forms on Mordell-Weil type groups.
Stefan Bara\'{n
In this paper we investigate an analogue of Hasse-Minkowski theorem for quadratic forms on Mordell-Weil type groups over number fields like $S$-units, abelian varieties with trivial ring of endomorphisms and odd algebraic $K$-theory groups.
1703.06089.tex

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\begin{document}





\title[Quadratic forms on  Mordell-Weil type groups]{Hasse-Minkowski theorem for quadratic forms on  Mordell-Weil type groups.}

\author{Stefan Bara\'{n}czuk} 
\address{Faculty of Mathematics and Computer Science, Adam Mickiewicz University\\ul. Umultowska 87, Pozna\'{n}, Poland}
\email{stefbar@amu.edu.pl}

\begin{abstract}In this paper we investigate an analogue of Hasse-Minkowski theorem for quadratic forms on Mordell-Weil type groups over number fields like $S$-units, abelian varieties with trivial ring of endomorphisms and odd algebraic $K$-theory groups.
\end{abstract}


\keywords{quadratic forms; Hasse-Minkowski theorem; Mordell-Weil groups} 

\subjclass[2010]{11R04; 11R70; 14K15} 


\maketitle



\section{Introduction.}
Let $V$ denote all places on $\mathbb{Q}$ including $\infty$. The Hasse-Minkowski theorem (cf. \cite{H}) states that a quadratic form over $\mathbb{Q}$ represents $0$ if and only if it represents $0$ 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.
%Thus we will have to limit our investigation to all but finitely many primes. 
Note however that the original Hasse-Minkowski theorem  also has its stronger versions where some number of places can be omitted (see Proposition \ref{2and3}); 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: 
\begin{enumerate}
	\item ${R}_{F,S}^{\times}$, $S$-units groups, where $F$ is a number field and $S$ is a finite set of ideals in the ring of integers ${R}_{F}$,
	\item $A(F)$, Mordell-Weil groups of abelian varieties over number fields $F$ with $\mathrm{End}_{\bar{F}} (A) = \mathbb{Z}$,
	\item $K_{2n+1}(F)$, $n>0$, odd algebraic $K$-theory groups.
\end{enumerate}	
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:\\
\begin{tabular}{ll}
$P \mod v$	&	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.						
\end{tabular}\\
We impose the following two assumptions on the family $r_{v} \colon B \to B_{v}$:\\

\noindent
\textbf{Assumptions.}
\begin{enumerate}
	\item \label{o posylaniu} 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 $\Z$ 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 $k_{i}=0$.
	\item \label{o torsyjnych} For almost all $v$ the map $B_{\mathrm{tors}} \to B_{v}$ is injective.
\end{enumerate}
(For the families of groups we described above the axioms are valid, in particular  
Assumption \ref {o posylaniu} is fulfilled by \cite{Bar1}, Theorem 5.1 and Assumption \ref{o torsyjnych} by \cite{BGK3}, Lemma 3.11. The validity of the technical hypothesis \footnote{note the misprint in its formulation: $\rho(G_{F})$ should read $\rho_{l}(G_{F})$} from the first sentence of Theorem 5.1 in \cite{Bar1} 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 \cite{B}.)\\

\section{Quadrartic forms of rank $2$.}

\begin{thm}\label{thm for 2}
 Let $P, Q\in B$ be points of infinite order. The following are equivalent:

\begin{itemize}
\item For almost every $v$ there exist coprime integers $x,y$ and a point $T \in B_{\mathrm{tors}}$ such that
\begin{equation}\label{zal dla 2}
x^2 P + y^2 Q  = T \mod v.
\end{equation}
\item
There exist coprime integers $x,y$ and a point $T \in B_{\mathrm{tors}}$ such that
\begin{equation}\label{teza dla 2}
x^2 P + y^2 Q  = T.
\end{equation}
\end{itemize}
\end{thm}

\begin{proof}
Suppose that $P,Q$ are linearly independent. Fix a prime number $l$ coprime to $\# B_{\mathrm{tors}}$. By Assumption \ref{o posylaniu} there is a positive density set of primes $v$ such that 
\[
\begin{array}{ccc}
l^{2} \parallel \mathrm{ord}_{v}P & \mathrm{and} & l^{3} \parallel \mathrm{ord}_{v} Q.
\end{array}
\]
Thus if both $x,y$ are coprime to $l$ then
\[
\begin{array}{ccc}
l^{2} \parallel \mathrm{ord}_{v}(x^2 P) & \mathrm{and} & l^{3} \parallel \mathrm{ord}_{v} (y^2 Q);
\end{array}
\]
if $l \mid x$ and $l \nmid y$ then
\[
\begin{array}{ccc}
l \nmid \mathrm{ord}_{v}(x^2 P) & \mathrm{and} & l^{3} \parallel \mathrm{ord}_{v} (y^2 Q);
\end{array}
\]
and if $l \nmid x$ and $l \mid y$ then
\[
\begin{array}{ccc}
l^{2} \parallel \mathrm{ord}_{v}(x^2 P) & \mathrm{and} & l^{2} \nmid \mathrm{ord}_{v} (y^2 Q);
\end{array}
\]
in all cases we get a contradiction to \eqref{zal dla 2}. Hence $P,Q$ are linearly dependent.\\

Write $aP+bQ=0$ for nonzero integers $a,b$. Multiplying \eqref{zal dla 2} by $b$ we get
\begin{equation}\label{tylko P}
(b x^2 - a y^2)P = b T \mod v.
\end{equation}
Fix an arbitrary prime number $l$ coprime to $\# B_{\mathrm{tors}}$. By Assumption \ref{o posylaniu} there is a positive density set of primes $v$ such that 
$l \parallel \mathrm{ord}_{v}P$. So by \eqref{tylko P} there are coprime integers $x,y$ such that  $l \mid (b x^2 - a y^2)$. Thus by Proposition \ref{2and3} \emph{\ref{P1}} the form $b x^2 - a y^2$ represents $0$. Putting $b x^2 - a y^2=0$ we get
$b(x^2 P + y^2 Q)  = 0.$




\end{proof}

\section{Quadrartic forms of rank $3$.}
 
\begin{thm}\label{thm for 3} Let $P, Q, R\in B$ be points of infinite order. The following are equivalent:

\begin{itemize}
\item 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
\begin{equation}\label{zal}
x^2 P + y^2 Q + z^2 R = T \mod v.
\end{equation}
\item
There exist integers $x,y,z$ with $\gcd(x,y,z)=1$ and a point $T \in B_{\mathrm{tors}}$ such that
\begin{equation}\label{teza}
x^2 P + y^2 Q + z^2 R = T.
\end{equation}
\end{itemize}
\end{thm}





\begin{proof}[Proof of Theorem \ref{thm for 3}]
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 \ref{o posylaniu} there is a positive density set of primes $v$ such that \begin{equation}\label{ord3 1}
\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}
\end{equation}
That gives
\begin{equation}\label{ord3 2}
\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}
\end{equation}
Now by \eqref{zal} we have $2 \mid z$ thus
\begin{equation}\label{ord3 4}
\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}
 \end{equation}
  so respectively 
\begin{equation}\label{ord3 3}
\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}
\end{equation}
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 \eqref{ord3 1} and \eqref{ord3 2} that 
\[ \begin{array}{ccc}
2^{1+e} \parallel \mathrm{ord}_{v} (y^2 (P+Q)) & \mathrm{and} & 2^{1+e} \nmid \mathrm{ord}_{v} ((x^2-y^2) P).
\end{array}\] 
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 \eqref{ord3 3}.   \\

If $x$ is odd and $y$ is even then 
\[\begin{array}{ccc}
2^{3+e} \parallel \mathrm{ord}_{v} (x^2 P) & \mathrm{and} & 2^{3+e} \nmid \mathrm{ord}_{v} (y^2 Q) 
\end{array}\]
but that together with \eqref{ord3 4} contradicts \eqref{zal}. 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\lbrace P, Q \right\rbrace $, $\left\lbrace P, R \right\rbrace $, $\left\lbrace Q, R \right\rbrace $ one is a pair of linearly dependent points. Without loss of generality we can assume that this pair is $\left\lbrace P, Q \right\rbrace $, i.e., there are nonzero rational integers $s,t$ such that $s P = t Q$. 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 \eqref{teza}. 

So suppose that $-s/t$ is not a square.
Multiplying \eqref{zal} by $s$ we get  
\begin{equation}\label{zal s}
(t x^2  + s y^2 ) Q +  s z^2 R = s T \mod v .
\end{equation}
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 \ref{o posylaniu}  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 \eqref{zal s} implies that $l \mid z$. This means that $l^2 \nmid \mathrm{ord}_{v} (s z^2 R)$  so again by \eqref{zal s} we get $l \mid (t x^2  + s y^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 \eqref{zal s} by $w$ we get  
\begin{equation}\label{zal s w}
( w t x^2  + w s y^2 +  u s z^2) Q  = w s T \mod v .
\end{equation}
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\lbrace e_{p} :  p \in \mathcal{P}  \right\rbrace $. Fix an arbitrary prime number $l$ and a positive integer $k$. By Assumption \ref{o posylaniu} there is a positive density set of primes $v$ such that $l^{k+e} \parallel \mathrm{ord}_{v} Q$ so $l^{k} \mid ( w t x^2  + w s y^2 +  u s z^2)$ by \eqref{zal s w}. Thus the quadratic form $( w t x^2  + w s y^2 +  u s z^2)$ represents 0 modulo arbitrary $l^{k}$ and hence by Proposition \ref{2and3} \emph{\ref{P2}} it represents 0.\\

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

Write
\begin{equation}\label{lin dep no pair}
aP+bQ+cR=0
\end{equation}
with $abc \ne 0$. Multiplying \eqref{zal} by  $a$ and \eqref{lin dep no pair} by $x^2$ and subtracting the results we get

\begin{equation}\label{lin dep QR}
(ay^2-bx^2)Q+(az^2-cx^2)R=a T \mod v.
\end{equation}
Fix a prime number $l$ coprime to the numbers $a,b,c,\# B_{\mathrm{tors}}$. By Assumption \ref{o posylaniu} there is a positive density set of primes $v$ such that 
\begin{equation}\label{PQorders}
\begin{array}{ccc}
l^{2} \parallel \mathrm{ord}_{v}Q & \mathrm{and} & l \parallel \mathrm{ord}_{v} R.
\end{array}
\end{equation}
By \eqref{lin dep QR} we get
\begin{equation}\label{abxy}
l \mid (ay^2-bx^2)
\end{equation} 
thus if $l$ divides either $x$ or $y$ then it divides them both.  But that implies by \eqref{PQorders} and \eqref{lin dep QR} that $l \nmid \mathrm{ord}_{v}(az^2R)$ 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 \eqref{abxy} 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 \ref{2and3} \emph{\ref{P1}} 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 \eqref{lin dep no pair} we get
\[a(sw)^2P+a(tw)^2Q+a(su)^2R=0.\] 
  
\end{proof}


\section{Quadratic forms of higher ranks.  Auxiliary results.}

Theorems \ref{thm for 2} and \ref{thm for 3} 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: 

\begin{prop}
Fix $n \ge 4$. Let $P$ be a point of infinite order. Define the form $f$ by
\begin{equation*}
( 2 x_{1}^2  + x_{2}^2  +x_{3}^2   +x_{4}^2  + \ldots + x_{n}^2)P.
\end{equation*}
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 
\begin{equation}\label{gt 4 mod v}
( 2 x_{1}^2  + x_{2}^2  +x_{3}^2   +x_{4}^2  + \ldots + x_{n}^2)P = 0 \mod v.
\end{equation}
\end{prop}

\begin{proof} The form $( 2 x_{1}^2  + x_{2}^2  +x_{3}^2   +x_{4}^2  + \ldots + x_{n}^2)$ is positive and the point $P$ is of infinite order so $( 2 x_{1}^2  + x_{2}^2  +x_{3}^2   +x_{4}^2  + \ldots + x_{n}^2)P = 0$ 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. \cite{Sie}, page 395, exercise 4) that $2k$ is of the form $2 a^2  + b^2  +c^2 +1$ where $a,b,c$ are integers. Thus in order to get \eqref{gt 4 mod v} we can put $x_{1}=a, x_{2}=b, x_{3}=c, x_{4}=1$ and $x_{i}=0$ for $i > 4$.
\end{proof}

 



\begin{prop}\hspace{1cm}\label{2and3}
\begin{enumerate}[label=\emph{(\alph*)}]
\item \label{P1}
Let $f$ be an integral quadratic form of rank $2$. If $f \mod v$ represents $0$ for all but finitely many $v\in V$ then $f$ represents $0$.  
\item \label{P2}
Let $f$ be an integral quadratic form of rank $3$. If $f$ represents $0$ in $\mathbb{Q}_{v}$ for all except at most one $v\in V$ then $f$ represents $0$.  
\end{enumerate}
\end{prop}



\begin{proof} 
\emph{\ref{P1}} This statement is an immediate corollary of the theorem proved in \cite{H2}.\\
\emph{\ref{P2}} Cf. \cite{S}, p. 43, Corollary 3.
\end{proof}
 

%-----------------BIBLIOGRAPHY----------


\bibliographystyle{plain}
\begin{thebibliography}{W}



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\bibitem[Bar]{Bar1} S. Bara\'{n}czuk, \textit{On reduction maps and support problem in $K$-theory and abelian varieties}, Journal of Number Theory 119 (2006), 1-17

\bibitem[Bog]{B} F.A. Bogomolov, \textit{Sur l'alg\'{e}bricit\'{e} des repr\'{e}sentations l-adiques},  
C. R. Acad. Sci. Paris Sér. A-B 290 (1980), no. 15, A701-A703

\bibitem[H1]{H} H. Hasse, \textit{\"{U}ber die Darstellbarkeit von Zahlen durch quadratische Formen im K\"{o}rper der rationalen Zahlen},
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\bibitem[H2]{H2} H. Hasse, \textit{Zwei Bemerkungen zu der Arbeit  \enquote{Zur Arithmetik der Polynome} von U. Wegner in den Mathematischen Annalen, Bd. 105, S. 628-631},  Math. Ann. 106 (1932), no. 1, 455-456

\bibitem[S]{S} J. -P. Serre, 
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\bibitem[Sie]{Sie} W. Sierpi\'{n}ski, 
\textit{Elementary Theory of Numbers}, North-Holland mathematical library, vol. 31, 1988 



\end{thebibliography}

\end{document}




1703.06089.extracted_bib
[BGK]{BGK3 G. Banaszak, W. Gajda, P. Kraso\'{n, Detecting linear dependence by reduction maps, Journal of Number Theory 115 (2005), 322-342
1703.06089.extracted_bib
[Bar]{Bar1 S. Bara\'{nczuk, On reduction maps and support problem in $K$-theory and abelian varieties, Journal of Number Theory 119 (2006), 1-17
1703.06089.extracted_bib
[Bog]{B F.A. Bogomolov, Sur l'alg\'{ebricit\'{e des repr\'{esentations l-adiques, C. R. Acad. Sci. Paris Sér. A-B 290 (1980), no. 15, A701-A703
1703.06089.extracted_bib
[H1]{H H. Hasse, \"{Uber die Darstellbarkeit von Zahlen durch quadratische Formen im K\"{orper der rationalen Zahlen, J. Reine Angew. Math. 152 (1923), 129-148
1703.06089.extracted_bib
[H2]{H2 H. Hasse, Zwei Bemerkungen zu der Arbeit \enquote{Zur Arithmetik der Polynome von U. Wegner in den Mathematischen Annalen, Bd. 105, S. 628-631, Math. Ann. 106 (1932), no. 1, 455-456
1703.06089.extracted_bib
[S]{S J. -P. Serre, A course in Arithmetic, Graduate Texts in Mathematics, Springer 1996
1703.06089.extracted_bib
[Sie]{Sie W. Sierpi\'{nski, Elementary Theory of Numbers, North-Holland mathematical library, vol. 31, 1988