most recent 30 from http://mathoverflow.net 2013-05-24T15:22:34Z http://mathoverflow.net/feeds/question/85435 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85435/perfectness-of-r-bilinear-form-preserved-under-extension-of-scalars Fabian Werner 2012-01-11T18:15:56Z 2012-01-11T18:27:07Z

<p>Hi all.</p> <p>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 $\Phi((a \otimes r), (b \otimes r')) = \phi(a, b) r r'$. </p> <p>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 $\mathbb{Z}$?</p> <p>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.</p> <p>The result was true if $A, B$ and $R$ were vector spaces over a field (see the reference in <a href="http://mathoverflow.net/questions/65802/optimal-reference-for-tensor-product-of-symmetric-bilinear-forms" rel="nofollow">http://mathoverflow.net/questions/65802/optimal-reference-for-tensor-product-of-symmetric-bilinear-forms</a>) but does this generalize to the module setting?</p> <p>Best and thanks,</p> <p>Fabian Werner</p>

http://mathoverflow.net/questions/85435/perfectness-of-r-bilinear-form-preserved-under-extension-of-scalars/85438#85438 Andreas Blass 2012-01-11T18:27:07Z 2012-01-11T18:27:07Z

<p>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.</p>