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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

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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.

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  • $\begingroup$ Aw, yes, this is certainly true... But what happens if none of them if finitely generated, does it still remain true? $\endgroup$
    – Fabian Werner
    Commented Jan 11, 2012 at 18:34
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    $\begingroup$ @Fabian : There are counterexamples : take $A=\prod_{n \in \mathbf{N}} \mathbf{Z}$ and $B=\bigoplus_{n \in \mathbf{N}} \mathbf{Z}$. The pairing $A \times B \to \mathbf{Z}$ given by $(a_n),(b_n)\mapsto \sum_{n \in \mathbf{N}} a_n b_n$ is perfect (surprising, but true, see here : math.lsa.umich.edu/~ablass/bis.pdf). After tensoring with $\mathbf{F}_p$ the pairing becomes $\prod_{n \in \mathbf{N}} \mathbf{F}_p \times \bigoplus_{n \in \mathbf{N}} \mathbf{F}_p \to \mathbf{F}_p$ which cannot be perfect because of dimension reasons. $\endgroup$
    – François Brunault
    Commented Jan 12, 2012 at 10:51
  • $\begingroup$ Since Francois cited a paper on which I'm a co-author, I should point out that this paper is not the original source for the result but merely quotes it. The perfectness of that pairing is, as far as I know, due to Ernst Specker in 1950. (Reinhold Baer had related results earlier, but I don't think he had this one.) Also note that this perfect pairing becomes imperfect when tensored with any field, for the same reason Francois gave. $\endgroup$
    – Andreas Blass
    Commented Jan 12, 2012 at 16:34
  • $\begingroup$ Thanks for the precisions. Also relevant is the previous MO post mathoverflow.net/questions/10239/… $\endgroup$
    – François Brunault
    Commented Jan 12, 2012 at 17:08
  • $\begingroup$ Hmm, i wonder what happens when they are general $R$-modules ($R$ not necessarily a PID), then one does not have the notion of a determinant because the Gramian matrix is not necessarily a square matrix, right? $\endgroup$
    – Fabian Werner
    Commented Jan 24, 2012 at 16:45

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