Let $M_\circ = \dots \to M_n \dots \to M_0 \to 0$ and $N_\circ = \dots \to N_n \dots \to N_0 \to 0$ be exact complexes of modules over a ring $A$ such that each module is flat.

Is it then true that $(M\otimes N)_\circ = \dots M_n\otimes_AN_n \dots \to M_0\otimes_A N_0 \to 0$ is exact?

I can get an exact double complex but I don't know how to use that to conclude what I want.

(The motivation is to show that if $P_\circ$ and $Q_\circ$ are polynomial simplicial resolutions of rings $B,C$ flat over $A$, then $P_\circ\otimes_AQ_\circ$ is a polynomial simplicial resolution of $B \otimes_A C$.

  • $\begingroup$ So, is the $n$-th term of the complex just $M_n\otimes N_n$? $\endgroup$ – Lord Shark the Unknown May 21 at 2:40
  • $\begingroup$ Yes. I am not actually sure if I need the tensor product associated to the total complex or this one but in the other case I know why it is exact. $\endgroup$ – Asvin May 21 at 2:41
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    $\begingroup$ I'd be amazed if this one were exact .... $\endgroup$ – Lord Shark the Unknown May 21 at 2:42

No. For instance, let both complexes be $0\to A\to A\to 0$ but with one of them in degrees $0$ and $1$ and the other in degrees $1$ and $2$. Then the tensor product will be nonzero only in degree $1$ and so will not be exact.

  • $\begingroup$ Do you know why the Eilenberg-Zilber theorem doesn't apply in this context? If I am reading it right, that says that the total double complex is quasi isomorphic to the diagonal complex in my question and the total double complex has trivial homology in this case. $\endgroup$ – Asvin May 21 at 3:24
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    $\begingroup$ The Eilenberg-Zilber theorem says that the total complex of the double complex is quasi-isomorphic to the diagonal complex of simplicial objects, not of chain complexes. In other words, you have to take the simplicial objects which correspond to the chain complexes under the Dold-Kan correspondence and take their diagonal tensor product. $\endgroup$ – Eric Wofsey May 21 at 3:44
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    $\begingroup$ This is analogous to how if you take two simplicial complexes and consider them as semisimplicial sets (with just the nondegenerate simplices), their product as semisimplicial sets does not correspond to the geometric product of realizations. But if you instead consider them as simplicial sets with their degenerate simplices, then the product is correct, the point being that you can get nondegenerate simplices in the product which are degenerate on each coordinate. $\endgroup$ – Eric Wofsey May 21 at 3:50
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    $\begingroup$ So, in the same way, when you turn a chain complex into a simplicial object, you adjoin degeneracies. When you then take a diagonal tensor product, those degeneracies actually matter because a tensor of two degenerate elements can be nondegenerate. $\endgroup$ – Eric Wofsey May 21 at 3:53
  • $\begingroup$ Thanks for the explanation! I think I knew that before and in this case my mistake was that the tensor product of the alternating sum of boundary maps is not the same as the alternating sum of the tensor products! $\endgroup$ – Asvin May 21 at 3:58

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