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 $I+J$.



when tensor complex resolves S/I+J? 
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.


