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EDIT : I edited the question according to Prof. Rickard's suggestions

Let $Y$ be an affine variety over $\mathbb{C}$ and $A$ and $B$ be $2$ algebras with finite homological dimension over $Y$ such that $D^b(Y,A)$ and $D^b(Y,B)$ are equivalent as $\mathbb{C}$-linear triangulated categories. Assume moreover that his equivalence is given by a two-sided tilting complex of $A^{op} \otimes^L_{\mathbb{C}} B$-modules, say $T$ (so that the equivalence between $D^b(Y,A)$ and $D^b(Y,B)$ is standard in the terminology of Rickard).

Let $M_A$ and $N_A$ be $2$ finitely generated $A$-modules and $\sigma : M_A \rightarrow N_A$ be a map of $A$-modules such that $\sigma$ does not vanish anywhere on $Y$. Assume finally that $M_A \otimes ^L_{A} T$ and $N_A \otimes^L_A T$ are pure $B$-modules (that is complexes of $B$-modules concentrated in degree $0$). Is it possible that the vanishing locus of the induced map: $$ \sigma \otimes^L_A T : M_A \otimes ^L_{A} T \rightarrow N_A \otimes^L_A T$$

is not empty?

Thanks a lot!

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I'm not sure exactly how to interpret your question. By an "algebra on" an affine variety $Y=\operatorname{Spec}(R)$, do you just mean an $R$-algebra? And by "bicomplex" I presume you mean "complex of bimodules", but do you mean $A^{op}\otimes_R B$-modules or $A^{op}\otimes_\mathbb{C}B$-modules? And what's a Rickard equivalence? – Jeremy Rickard May 10 '13 at 20:44
@ Jeremy Rickard : does the answer depend on $T$ being a $A^{op} \otimes_R B$-module or a $A^{op} \otimes_{\mathbb{C}} B$-module ? – Johan May 10 '13 at 21:50

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