Let $A$ be a commutative ring, $M$ and $N$ $A$-modules. Is the derived tensor product $M[0]\otimes^L N[0]$ isomorphic to $M\otimes_A N$? I know that the derived tensor product is supposed to be a "redefinition" of the usual tensor product.
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Welcome to Math.SE! Could you tell us with which definition of the derived tensor product you are working? – Hanno Jul 19 '16 at 05:12
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@Hanno the derived tensor product is the right derived functor of the tensor product functor (like in Gelfand/Manin Hom' algebra). Is there another definition? – Richard Pink Jul 19 '16 at 22:39
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No it is not the same thing, but you can say that $H_0(M[0]\otimes^L_A N[0])=M\otimes_A N$. In fact, $H_i(M[0]\otimes^L_A N[0])=\operatorname{Tor}^A_i(M,N)$. – Roland Jul 20 '16 at 15:16
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In general, no. As Roland notes in his comment, $$ H_i(M\otimes^\mathbf{L}N) = \text{Tor}_i^A(M,N) $$ When you first construct tor, you take a flat resolution $F_\bullet \to M$ or $G_\bullet \to N$ and compute the homology of the complex $F_\bullet\otimes N$ or $M\otimes G_\bullet$. This situation is mirrored in the derived category setting where $M\otimes^\mathbf{L}N$ is quasi-isomorphic to $F_\bullet \otimes G_\bullet$, $F_\bullet\otimes N$, or $M \otimes G_\bullet$. This follows from a spectral sequence/double complex type of argument. Hence you only get a quasi-isomorphism of $M\otimes^\mathbf{L}N$ with $M\otimes N$ if and only if $M$ or $N$ is flat.
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