Hi all.

I have the following setting: $A, B$ are $\mathbb{Z}$-modules (in my case, $B$ is free and finitely generated) and i have a $\mathbb{Z}$-bilinear map $\phi:A \times B \mapsto \mathbb{Z}$. Now i want to do an "extension" of scalars, meaning that i take an arbitrary commutative ring $R$ with unit ($\mathbb{F}_p$ for example), then there is a unique $R$-bilinear form $\Phi : (A \otimes R) \times (B \otimes R) \mapsto \mathbb{Z} \otimes R$ satisfying $\Phi((a \otimes r), (b \otimes r')) = \phi(a, b) r r'$.

The question now is: if $\phi$ is perfect (meaning that the maps $A \mapsto d_{\mathbb{Z}}(B), a \mapsto \phi(a, \cdot)$ and $B \mapsto d_{\mathbb{Z}}(A), b \mapsto \phi(\cdot, b)$ where $d_R(\cdot) = Hom_{R}(\cdot, R)$ are bijective), is this also true for $\Phi$ with $R$ in place of $\mathbb{Z}$?

I already know that non-degeneracy alone is not sufficient, if we take a lattice $L = \mathbb{Z} x$ and the bilinear form $\phi(x,x) = p$ and then tensor with $\mathbb{F}_p$, this gives $\Phi(x \otimes \overline{1}, x \otimes \overline{r}) = p = 0 \mod p$ so that $\Phi$ is degenerate.

The result was true if $A, B$ and $R$ were vector spaces over a field (see the reference in Optimal reference for tensor product of symmetric bilinear forms?) but does this generalize to the module setting?

Best and thanks,

Fabian Werner