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 $B \otimes_A C$.

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.
