Prime ideals of a finite direct product ring 
Is it true that any prime ideal of a finite direct product ring $R=\prod_{i=1}^nR_i$ is of the form $P=\prod_{i=1}^nI_i$, where $I_j$ is a prime ideal of $R_j$ for some $j$ and $I_i=R_i$, for $i\neq j$?

Any ideal of the form above is prime in $R$. Indeed, if $X=\prod_{i=1}^nA_i$ and $Y=\prod_{i=1}^nB_i$ are ideals of $R$ (where $A_i$'s and $B_i$'s are ideals of $R_i$) such that $XY\subseteq P$, then either $A_j$ or $B_j$ is a subset of a  prime ideal $I_j$, for some $j$. Hence either $X$ or $Y$ is a subset of $P$.
Any help/suggestion would be appreciated!
 A: If I am not mistaken, you've proven ideals of that form are prime, and now you are headed for the converse. Let $I$ be a prime ideal of $R$.
I assume you already know the ideal structure of finite products of rings, so you can see why the ideal must be of the form $I=\prod I_i$ for some ideals $I_i\lhd R_i$. Then $R/I=(\prod R_i)/(\prod I_i)\cong\prod R_i/I_i$.
Now, in order for this product to be a prime ring (which is exactly what a quotient by a prime ideal gives you) it is necessary for all factors to be zero except one, and the one that is nonzero must also be a prime ring.
The first condition implies there is a $j$ such that $R_k=I_k$ for all $k\neq j$, and the second condtion implies $I_j$ is a prime ideal of $R_j$.

A quick note about KonstantinArdakov's comment-solution: it's fine, except for a small danger of readers misunderstanding part of the logic.
As we know, the general definition of prime ideals says that $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$ for any two ideals $A,B$ of $R$. That is, it's the ideal-wise version of the element-wise condition for primality. So something needs to be said about why $e_ie_j=0$ implies one of the two is in $P$.
Now, given that at least one of $a$ and $b$ is central, we can conclude that $(a)(b)=\{0\}\subseteq P$, and it follows that $a\in P$ or $b\in P$, so everything works as hoped.
