when tensor complex resolves S/I+J?  Assume that $I\subset k[x_1,\ldots,x_n]$ and $J\subset k[y_1,\ldots,y_m]$ are monomial ideals in different rings, and the minimal free resolution of $S/I$ and $S/J$, say $F_\cdot$ and $G_\cdot$, are both linear. I believe that $F\otimes G$ is a minimal free resolution for $S/I+J$. Does anyone have any comment for the proof, or any reference?  
The tensor product of two chain complexes $(A,d_1)$ and $(B,d_2)$, say $A\oplus B$, is formed by taking all products 
$A_i \otimes B_j$ and letting $(A \otimes B)_k$ be the direst some of $A_i\otimes B_j$ for $i+j=k$. The differential maps are  defined as $\partial(a\otimes b) = d_1a \otimes b + (-1)^i a \otimes d_2b$ when $a\in A_i$. Then we have $\partial^2 = 0$.
So in order to prove the question we need to check the exactness of the complex, and how it resolves the resolution of $I+J$.
 A: There are really two separate things being asked.  (1) When is the complex $F\otimes G$ exact?  (2) If it is exact, when is $F\otimes G$ a minimal free resolution?
The first question is computed by Tor.  Namely $F\otimes G$ is exact if and only if $\text{Tor}_i(S/I,S/J)=0$ for all $i>0$
I believe that the second question is easier.  Since the differential $\partial$ on $F\otimes G$ is defined in terms of differentials on $F$ and $G$ (which were assumed to be minimal free resolutions), we see that $\partial (F\otimes G)_i$ belongs to the maximal ideal times $(F\otimes G)_{i-1}$.  Thus, $F\otimes G$ is a minimal free resolution if and only if it is exact.
Of course, in your example where $S=k[x_1,\dots,x_n,y_1,\dots,y_m]$, and $I$ only involves $x$-variables and $J$ and only involves $y$-variables, then the higher Tor's vanish and thus $F\otimes G$ is a minimal free resolution.
A: I'll use Daniel's notation.
One way to see that $F\otimes G$ is exact is to notice the following. If $A$ and $B$ are the polynomial rings generated by the $x$s and the $y$s, respectively, then the minimal resolutions of $S/I$ and $S/J$ can be constructed first constructing the minimal resolutions $\bar F$ and $\bar G$ of $A/I$ and $B/J$ and then extending scalars to $S$. Therefore $F\otimes_S G$ can in fact be gotten as $\bar F\otimes_k\bar G$ as a complex of modules over $S=A\otimes_k B$. Now the Künneth formula tells  us that the tensor product of two exact complexes of $k$-vector spaces is exact, and this applies to $\bar F\otimes_k\bar G$.
