Reference request: Given a non-degenerate integral quadratic lattice $L,q$ over a PID, the quotient $L^*/L$ is given by SNF of $q$

Question

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!

Answer 1

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.