Let $A$ be a commutative ring with $1$ and $M,N$ be $A$-modules. Can you give a quick proof that $\textrm{Tor}_i(M,N) \cong \textrm{Tor}_i(N, M)$ using derived categories?

In his Homological algebra book, Weibel proves this with an argument via a double complexes: the so-called "acyclic assembly lemma", and from what I understand this argument can be essentially reworded into the language of spectral sequences. Hartshorne's discussion (in "Residues and Duality") of derivatives of functors in two variables is quite short, but it's not clear to me if this result (commutativity of Tor) immediately follows from the relevant derived category formalism.

  • $\begingroup$ No derived categories, but an explanation of what Weibel does. With double complexes he proves the balancing, i.e. that Tor can be calculated by deriving either M⊗− or −⊗N. This can be done without double complexes, by showing that one of these is a universal δ-functor in the other argument as well (where it is not derived). This is the proof suggested in the Tohoku paper. The commutativity of Tor for commutative rings is proved (in the language of universal δ-functors) by Weibel in the remark preceding Corollary 3.1.5.: he notes that one derives isomorphic functors M⊗− and −⊗M. $\endgroup$ – user2171 Sep 22 '16 at 0:22

Yes, the derived category of a commutative ring is symmetric monoidal $M\otimes_A^{\mathbb L}N=N\otimes_A^{\mathbb L}M$ and Tor is the homology of this tensor product.

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    $\begingroup$ To relate this to the usual definition of Tor (derived functors of $M\otimes -$ or $-\otimes N$), presumably you need to show that the derived tensor product $M\otimes^L N$ can be computed by resolving only one of the two factors. Doesn't that require some kind of double-complex type argument? $\endgroup$ – Charles Rezk Apr 9 '13 at 14:22
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    $\begingroup$ You only need to know that cofibrant objects are flat, i.e. tensoring with a cofibrant object preserves weak equivalences. $\endgroup$ – Fernando Muro Apr 10 '13 at 21:04

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