# Tensor product of exact complexes is exact

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 $$B \otimes_A C$$.

• So, is the $n$-th term of the complex just $M_n\otimes N_n$? – Lord Shark the Unknown May 21 at 2:40
• 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. – Asvin May 21 at 2:41
• I'd be amazed if this one were exact .... – 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.