TITLE: How does Tor and the Tensor functor interact?
QUESTION [5 upvotes]: So I've run into this question while doing some computations and I'm unsure if what I'm trying to show is true. Assume tensors are over $\mathbb{Z}$, is
$\mathrm{Tor}(A,\mathrm{Tor}(B,\mathbb{Z}/p\mathbb{Z})) = \mathrm{Tor}(A,\mathbb{Z}/p\mathbb{Z}) \otimes \mathrm{Tor}(B,\mathbb{Z}/p\mathbb{Z})$
and 
$\mathbb{Z}/p\mathbb{Z} \otimes \mathrm{Tor}(B,\mathbb{Z}/p\mathbb{Z}) = \mathrm{Tor}(B,\mathbb{Z}/p\mathbb{Z})$
In otherwords, does Tor absorb tensors with 'reasonably friendly' $\mathbb{Z}$-modules. If not can a simple counter example be found?

REPLY [5 votes]: The second statement is true just because $Tor(B,\mathbb Z/p)$ is annihilated by $p$ (since one its arguments is), and a $\mathbb Z$-module annihilated by $p$ is the same thing as a $\mathbb Z/p$-module, which then doesn't change when you tensor with $\mathbb Z/p$. 
As for the first statement, as already noted, $Tor(B,\mathbb Z/p)$ is a $\mathbb Z/p$-module, and so we can ask more generally if $M$ is a $\mathbb Z/p$-module,
is there an isomorphism $Tor(A,M) \cong Tor(A,\mathbb Z/p) \otimes M?$  Now a module over $\mathbb Z/p$ is just a vector space, so is a direct sum of copies of $\mathbb Z/p$.  Since $Tor$ commutes with direct sums,  there is indeed such an isomorphism; thus the first statement is also true.