Let $V$ be a finitely generated module over the ring $R=K\times K$ where $K$ is a field. We fix the switch involution on the ring $R$. Let $H$ be a hermitian form over $V$.

When $V$ is a free module, $H$ will be given by a matrix from $M_n(K\times K)\cong M_n(K)\times M_n(K)$ where $n$ is the rank of $V$. Further the hermitian condition gives that it should look like $(A,A^t)$ and using this I can prove that there is a unique hermitian form up to equivalence. I feel that this should be true in general, that is, I don't need to assume free. Can someone please provide any reference for this? Thanks in advance.