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?](https://mathoverflow.net/questions/65802/optimal-reference-for-tensor-product-of-symmetric-bilinear-forms)) but does this generalize to the module setting?


Best and thanks,


Fabian Werner



Perfectness of R-Bilinear form preserved under extension of scalars?
As long as your $B$ is free and finitely generated, $A$, being isomorphic to $d\_{\mathbb Z}(B)$, is also free on the same number of generators. Fix free generators $a\_i$ for $A$ and $b\_i$ for $B$. Perfectness of $\phi$ will make the matrix with entries $\phi(a\_i,b\_j)$ have determinant 1. That determinant will remain 1 after you extend scalars, and so $\phi$ will remain perfect.


