S genus of quadratic forms

Question

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.

Answer 1

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.