Equivalence of quadratic forms over $p$-adic integers vs over localisation at $p$

Question

To discern whether two integral quadratic forms are equivalent over the $p$-adic integers, one can compute a Jordan decomposition at $p$ and read off some invariants. Restricting to $p\ne2$ for convenience, Conway and Sloane give an algorithm for this procedure (`$p$-adic diagonalisation'). But their algorithm in fact computes a diagonalisation over the localisation of $\mathbb{Z}$ at $p$, $\mathbb{Z}_{(p)}:=\{\frac{a}{b}\text{ : }a,b\in\mathbb{Z},p\nmid b\}$, rather than over the $p$-adic integers $\mathbb{Z}_p$ in generality. My question is: does equivalence over $\mathbb{Z}_{(p)}$ differ from equivalence over $\mathbb{Z}_p$? And if so, when do such differences occur?

Answer 1

Since $\mathbb{Z}_{(p)}$ can be thought of as a subring of $\mathbb{Z}_{p}$, if two quadratic forms $Q_{1}$ and $Q_{2}$ are equivalent over $\mathbb{Z}_{(p)}$, then they must be equivalence over $\mathbb{Z}_{p}$. (And this is the reason that it suffices for Conway and Sloane to compute a diagonalization over $\mathbb{Z}_{(p)}$.)

However, the converse is false. Equivalence over $\mathbb{Z}_{p}$ does not imply equivalence over $\mathbb{Z}_{(p)}$ and maybe a good way of explaining this is that $\mathbb{Z}_{p}$ has a lot fewer square classes. In particular, if $p > 2$ and $x \in \mathbb{Z}_{p}$, then $x$ is a square if and only if $x = p^{2k} u$ for some integer $k \geq 0$ and $u \in \mathbb{Z}_{p}^{\times}$ whose reduction $\tilde{u} \in \mathbb{F}_{p}$ is a square.

In particular, the quadratic forms $Q_{1}(x,y) = x^{2} - y^{2}$ and $Q_{2}(x,y) = x^{2} - 7y^{2}$ are equivalent in $\mathbb{Z}_{3}$, but are not equivalent in $\mathbb{Z}_{(3)}$. This follows from the observation that $\sqrt{7} \in \mathbb{Z}_{3}$, while $\sqrt{7} \not\in \mathbb{Z}_{(3)}$.