Non-degenerate bilinear forms and orthogonal bases for free modules

Question

Given a finite-dimensional free module $M$ over a (commutative) ring $R$, let $(-,-)$ be a symmetric non-degenerate $R$-bilinear map $M \times M \to R$. Will $M$ admit two $R$-module bases, $e_i$ and $e_j$, such that $$ (e_i,e_j) = c_{ij} \delta_{ij}, ~~ \textrm{ for some } c_{ij} \in R? $$ To clarify, by non-degeneracy I mean that for any $m \in M$, $\exists$ an $n \in M$ s.t $(m,n) \neq 0$.

Answer 1

No, for example take $R=\mathbb{Z}$ and $M=\mathbb{Z}$, and the form $(m,n)=2mn$. The image of this form is $2\mathbb{Z}$, and so does not contain $1$, thus a basis as you require does not exist.

This can be generalized to any ring which has a non-invertible element $a$ which is not a zero divisor. Take $M=R$ and the non-degenerate form $(r,s)=ars$.

EDIT: The question can be phrase as is whether any symmetric (non-degenerate) matrix is equivalent to a diagonal matrix. This is not true, for example, take the ring $\mathbb{Z}[x]$ and the matrix $A=\pmatrix{x&2\\2&x}$. If $A$ would be equivalent to a diagonal matrix $D=\pmatrix{d_1 &\\&d_2}$ then they will have the same determinant (up to $\pm$, so we can assume the determinants are actually the same) and their entries will generate the same ideal (see here).

Combine this we get $d_1d_2=x^2-4$ and $\langle d_1,d_2\rangle=\langle x,2\rangle$. This is impossible, by direct calculation using $d_1=ax+2b$,$d_2=qx+2r$ we get that (up to $\pm$ and switching $d_1$ and $d_2$) $d_1=x+2$ and $d_2=x-2$. The last contradict the assumption about the ideals.

I hope this solve the question, and sorry again about the mistake.