Projective modules over a direct product of rings 
Let $R$ and $S$ be rings, and let $\text{proj}(R)$ denote the category of finitely generated projective modules. Is there an equivalence of categories between $\text{proj}(R \times S)$ and $\text{proj}(R) \times \text{proj}(S)$?

So far, I've managed to convince myself that if $P$ is a projective $R$-module and $Q$ is a projective $Q$ module, then $P \times Q$ (where $R \times S$ acts via $(r,s) \cdot (p,q) = (rp,sq)$) is a projective $(R \times S)$-module. Indeed, we have $P \oplus P' \cong R^n$, $Q \oplus Q' \cong S^n$ (here, without loss of generality, we assume that the number of summands of $R$ and of $S$ are the same) and so
$$ (P \times Q) \oplus (P' \times Q') \cong (P \oplus P') \times (Q \oplus Q') \cong R^n \times S^n \cong (R \times S)^n.$$
However, I can't convince myself that a projective $(R \times S)$-module is  a product of projective $R$ and $S$ modules. I don't even know if this is true. I'm wondering if we require our categories to be Krull-Schmidt.
 A: A module over $R\times S$ can be seen as an ordered pair $(M,N)$, where $M$ is an $R$-module and $N$ is an $S$-module. It's quite easy to see that $(M,N)$ is projective if and only if both $M$ and $N$ are projective.
First of all, if $L$ is a (left) module over $R\times S$ we can consider $M=(1,0)L$ and $N=(0,1)L$. Then, as groups, $L=M\oplus N$; moreover $M$ is an $R$-module via $rx=(r,0)x$ and similarly for $N$.
Conversely, given a pair $(M,N)$ we can define a structure of module over $R\times S$ on $M\times N$ by $(r,s)(x,y)=(rx,sy)$. The module resulting from the pair $((1,0)L,(0,1)L)$ is (isomorphic to) $L$.
The direct sum of two pairs $(M_1,N_1)$ and $(M_2,N_2)$ is $(M_1\oplus M_2,N_1\oplus N_2)$ (check it).
Since the pair corresponding to $(R\times S)^n$ is clearly $(R^n,S^n)$, we see that the pair corresponding to a finitely generated projective module over $R\times S$ consists of a finitely generated projective $R$-module and a finitely generated projective $S$-module.
Conversely, suppose $M$ is a direct summand of $R^m$ and $N$ is a direct summand of $S^n$. Then they can be seen as direct summands of $R^{m+n}$ and $S^{m+n}$ respectively. This makes $(M,N)$ into a direct summand of a free module over $R\times S$.
