Let $R$ be a PID with field of fraction $K$.
Let $L$ be a lattice with non-degenerate quadratic form $q:L\times L \to R$.
Let
$$
L^\* = \{x \in L\otimes K \text{ s.t. } q(x,l) \in R \text{ for all } l \in L \}.
$$
By integrality of $q$, we have $L \subseteq L^\*$.
I heard the following


**Claim.** The unique decomposition of the quotient $L^\*/L$ given by the structure theorem of modules over principal ring is exactly the one given by the Smith normal form of $q$.


I think it should pretty much follow from the definitions, but I'm a bit confused about how the argument goes. I checked OMeara's and Cassel's books but without success. I'd very much welcome a reference.
Thanks!



Reference request: Given a non-degenerate integral quadratic lattice $L,q$ over a PID, the quotient $L^*/L$ is given by SNF of $q$
The following more general statement is easier to prove:


Let $L\_1, L\_2$ be lattices with a nondegenerate bilinear form $b: L\_1 \times L\_2 \to R$. Let $$L\_1^\* = \{ x \in L\_2 \otimes K \textrm{ s.t. } b(l,x) \in R \textrm{ for all } l \in L\_1 \}$$


Claim: The unique decomposition of the quotient $L\_1^\*/L\_2$ given by the structure theorem of modules over a principal ring is exactly the one given by the Smith normal form of $b$.


Proof: Putting a matrix in Smith normal form involves acting by invertible matrices on the left and the right, which corresponds to invertible changes of coordinates in $L\_1$ and $L\_2$, which do not affect the quotient $L\_1^\* /L\_2$. So we may assume that $b$ is already in Smith normal form - in particular, is a diagonal matrix with diagonal entries $a\_1,\dots, a\_n$. Then if $e\_1,\dots, e\_n$ is a basis for $L\_2$, $a\_1^{-1} e\_1,\dots, a\_n^{-1} e\_n$ is a basis for $L\_1^\*$, and so $$L\_1^\*/L\_2 = \prod\_i a\_i^{-1} e\_iR /e\_iR =\prod\_i R/a\_i,$$ as desired.


