I'm reading Lei Fu's "Etale Cohomology Theory".
How to show this last condition is equivalent to saying the bilinear form in the proposition is nondegenerate?
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One says (I guess) a bilinear form on a locally free $\mathcal O_S$-module of finite rank $\mathcal A$ is nondegenerate if the induced morphism $\mathcal A \to \mathcal{Hom}_{\mathcal O_S}( \mathcal A, \mathcal O_S)$ is an isomorphism.
Locally an an open set where $\mathcal A$ is free, this is equivalent to the determinant of the map (well-defined after choosing a basis) being invertible. Now use the fact that an element of a ring is invertible if and only if it is invertible modulo every prime ideal to see that nondegeneracy is equivalent to nondegeneracy modulo every prime ideal.
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$\begingroup$ Can we deduce $A_{\mathfrak q}\times A_{\mathfrak q}\to B_{\mathfrak q}$ is nondegenerate from $A_{\mathfrak q}/\mathfrak qA_{\mathfrak q}\times A_{\mathfrak q}/\mathfrak qA_{\mathfrak q}\to B_{\mathfrak q}/\mathfrak qB_{\mathfrak q}$ is nondegenerate? $\endgroup$ Commented May 31 at 17:41
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$\begingroup$ @Borntobeproud Yes, because an element in a local ring being invertible is equivalent to its reduction modulo the maximal ideal being invertible. $\endgroup$ Commented May 31 at 18:38
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$\begingroup$ But there is another definition of nondegeneracy of $\mathcal A\times\mathcal A\to \mathcal O_S$, that is, for any open set $U\subseteq S$, if there exists some $x\in \mathcal A(U)$ such that for all $y\in \mathcal A(U)$, we have $\operatorname{Tr}(xy)=0$, then $x=0$. Are these two definitions of nondegeneracy of $\mathcal A\times\mathcal A\to \mathcal O_S$ equivalent in the proposition? $\endgroup$ Commented Jun 1 at 7:17
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$\begingroup$ @Borntobeproud No. A ramified covering of curves, i.e. $f\colon \mathbb A^1 \to \mathbb A^1, t\mapsto t^2$, will give a counterexample. The pairing $\operatorname{Tr}(xy)$ takes two polynomials in $t$ (or rational functions, on an open set) to their product after all odd degree terms is removed. As a bilinear form on the free module over $k[t^2]$ with basis $1,t$, this is given by the matrix $\begin{pmatrix} 1 & 0 \\ 0 & t^2 \end{pmatrix}$. It's easy to see from this that the bilinear form is nondegenerate in your sense but not mine. $\endgroup$ Commented Jun 1 at 10:03