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Let $f$ be a non-degenerate quadratic form with integral coefficients. The genus of $f$ is the set of quadratic forms up to integral equivalence which are equivalent to $f$ over the $p$-adic integers $\mathbb Z_p$ for all primes $p$ and also over $\mathbb Q$. Let us denote it by gen(f).

A $S$-arithmetic analog is the following: Let $S$ be a non-empty subset of primes. Let $gen_S(f)$ denote the set of all quadratic forms up to integral equivalence which are equivalent to $f$ over $\mathbb Z_p$ for all primes $p\in S$ and over $\mathbb Z_S$, where $\mathbb Z_S=\mathbb Z[\{\frac{1}{p}:p\in S\}]$.

It is well known that $gen(f)$ (and therefore also $gen_S(f)$) are finite sets.

It is clear to me that there exist a finite set $S$ with $gen_S(f)=gen(f)$ (as any $\mathbb Q$-isomorphism is defined over $\mathbb Z_S$ for some $S$), but I'm interested in an opposite case:

Assume that $|gen(f)|\neq 1$. Is there a set (non-empty as above) $S=S(f)$ for which $gen_S(f)$ has size one?

Please mention any reference which you think is relevant.

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never heard of it. This seems to belong to an existing area of study; what are some references on your S-arithmetic? –  Will Jagy Dec 1 '13 at 21:44
    
I have no references! but it seems very natural to me. If you think on the class group of $SO(f)$ (which is $SO(Q)\setminus SO(A)/SO(R\times \prod Z_P)$ then it correspond to the regular notion of genus. If you do the same thing but no with all the adeles, but just the completions at $S$ you'll get the above notion. One the aims of this questions is to get references...(reference for class groups of algebraic groups platonov rapinchuk book chapter 9). –  Menny Dec 1 '13 at 22:01
    
Sounds good. I was surprised because I and Alex Berkovich published an article in which we defined an $S$-genus, a collection of genera of positive integral ternary quadratic forms. Alex began with `supergenus' or similar name, but that was already used for something else. Well, there are at least two experts who look at MO intermittently. I will inform at least one of them about this question. –  Will Jagy Dec 1 '13 at 22:31
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@Menny: You meant to include "equivalent to $f$" in some places in your definitions (otherwise you aren't using $f$), to use orthogonal groups rather than special orthogonal groups, and to assume $f$ is non-degenerate over $\mathbf{Q}$. Your ${\rm{gen}}_S(f)$ is the kernel (in the sense of pointed sets) of the map ${\rm{gen}}(f) \rightarrow {\rm{gen}}(f_S)$, where $f_S$ denotes the associated quadratic module over $\mathbf{Z}_S$ (over which there is the evident notion of "genus" as over any Dedekind domain). Clearly $S=\emptyset$ answers your question, so did you want $S$ not empty? –  user76758 Dec 2 '13 at 2:23
    
@user76758 - thanks. I agree with your comments and it is indeed good to note that this notion exists over any Dedekind domain. I did not forget to exclude the $S=\emptyset$ but I added it also in the question to clarity. I did not specify what I mean by equivalent - should I restrict it to SO(q) for some reason? –  Menny Dec 2 '13 at 8:03

1 Answer 1

up vote 4 down vote accepted

Let $L=(\mathbf Z^n,b)$ be a bilinear module such that $L\otimes \mathbf Q$ is non-degenerate.

For a set of ultrametric places $S$, let us write $\mathbf Z[S^{-1}]$ for the set of rationals that are integral at each place out of $S$, and $\mathbf Z_{(S)}$ for the set of rationals that are integral at each place in $S$, so that $\mathbf Z[S^{-1}]$ is the same as $\mathbf Z_{(T)}$ when $T$ is the complement of $S$.

Let $\mathrm{Gen}(L)$ be the genus of $L$. One may ask two distinct questions :

(Q1) Does there exist a finite set of ultrametric places $T$ such that for all $M$ in $\mathrm{Gen}(L)$, there exists an isomorphism from $L\otimes \mathbf Z[T^{-1}]$ to $M\otimes \mathbf Z[T^{-1}]$ ?

(Q2) What is the biggest subset $S$ of the set of ultrametric places such that for all $M$ in $\mathrm{Gen}(L)$, one has an equivalence between the two affirmations $L\otimes \mathbf Z[S^{-1}]\simeq M\otimes \mathbf Z[S^{-1}]$ and $L\simeq M$.

In my first answer, I assumed you were asking for (Q1). I am not sure of this anymore. So here are answers to both questions.

(Q1) : This is an application of the strong approximation theorem.

Recall that it says that if $T$ is a finite set of places containing the archimedean ones, and if $S$ is its complement in the set of all places, and if $L$ is isotropic at at least one place in $T$, then the spinor genus of $L\otimes \mathbf Z_{(S)}$ contains a single class. Now at least when the dimension exceeds $3$, the set of ultrametric places $p$ where the equivalence relation defining spinor genera differs from the equivalence relation defining genera is finite (the two equivalence relations at the place $p$ agree when the determinant is a unit in $\mathbf Z_p$). Thus it suffices to choose $T$ big enough to contain all of these "bad" classes, and also a place where the form is isotropic.

(Q2) : This biggest set can be empty. Here is an example : take for $(L,q)$ the sum $I_9$ of nine copies of $I_1:=(\mathbf Z,(x,y)\mapsto xy)$. There are two classes in the genus of $L$; the one is represented by $L$, the other is represented by $M:=\mathrm{E}_8\oplus I_1$. It happens that $M\otimes\mathbf Z[\frac{1}{p}]$ is isomorphic to $L\otimes\mathbf Z[\frac{1}{p}]$ for all prime $p$.

It can also be non-empty : For $I_n$ with $n\leq 8$, the set $S$ is the set of all ultrametric places, since then the genus of $I_n$ contains a single class.

These two cases are representative of the general situation : if the genus of $L$ contains a unique class, then $S$ consists of all places. If the spinor genus contains more than one class, then $S$ cannot contain any place where $L$ is isotropic, and thus is empty when the dimension exceeds 5.

The canonical reference for this kind of questions is O'Meara.

Addendum (4/12/13): In dimensions $2$, funny things happen : by a theorem of Gauss, the theory (at least on the level of proper genera) is the same as the theory of narrow class groups $C(O)$ of orders in quadratic extensions of $\mathbf Q$. That question $(Q1)$ admits a positive answer then follows from the fact that these class groups are finitely generated (and in fact even finite). And $(Q_2)$ asks for an order $O$ what is the set $S$ of primes $p$ such that the ideals dividing $pO$ are trivial in $C(O)$. I guess that the proportion of these primes in $S$ in $\frac{1}{\vert C(O) \vert}$ but might be wrong on this point.

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Thanks you, I probably didn't state clearly what I denote by $Z_S$. For example for $S=\{p\}$ it is $\mathbb Z[\frac{1}{p}]$ and for $S$ equal to all the primes it is $\mathbb Q$. In think that in your definition you just take the complement of $S$, so maybe you do answer the question - I still don't understand it fully. Can you get the result for $S$ that contains only one prime $p$ (using my definition)? Thanks! –  Menny Dec 2 '13 at 8:13
    
Thanks again. I meant Q2. For Q1 though, I don't see what I miss in the comment I made in the question - any element of genus is equivalent over $\mathbb Q$ and therefore (as any matrix has finitely many denominators) over $\mathbb Z[Sˆ{-1}]$ for some $S$. As the genus is finite, it follows that there is a finite $S$ for which any element of the genus of $f$ is equivalent to $f$ over $[Sˆ{-1}]$, which answers Q1. I still have questions about your solution to Q2, but I'm working on that... Thanks again.Menny –  Menny Dec 2 '13 at 20:55
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For (Q1), of course as you say, you don't need to use such a powerfull tool as the strong approximation theorem (SAT) to get the answer, but using it, you know exactly what a minimal solution S is. Moreover, the other affirmations I proposed as examples for (Q2) are also consequences of this theorem, but can be proved without it by naïve computations. My inclination is to use the SAT as soon as possible : it gives insight and precise results. –  few_reps Dec 2 '13 at 21:51
    
@Menny : so ... are you convinced now ? –  few_reps Dec 9 '13 at 0:15
    
More than convinced... I'm still browsing through O'Meara to fully understand. Hopefully I'll post references to your claim soon... Thanks a lot for your answer! –  Menny Dec 9 '13 at 22:51

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