Finite product of semilocal rings is semilocal

Question

We know a ring is semilocal if it has finitely many maximal ideals.

Prove that finite product of semilocal rings is semilocal.

I am trying to prove it but, so far can't having any luck. I am thinking about using the fact is if $R$ is semilocal ring then $R/J(R)$ is product of finite field. Where $J(R)$ is Jacobson radical of the ring. Any hint or help will be appreciated. Thanks in advance.

One thing I want to say, is this question is pretty easy if we consider the product of our rings to be direct product that is of the form $R_1×R_2$. But what I am having trouble understanding is what if the product they meant in the question is $R_1R_2$ where every elements is of the form $\sum x_iy_i$, with $x_i \in R_1, y_i \in R_2$. So it will be great if someone clarify this topic also.

Answer 1

Consider product of two rings, $ R=R_1×R_2$. Then every prime ideal in $ R$ is of the form $R_1×q$ or $p× R_2$, where $p$ and $q$ are prime ideal of $ R_1$ and $R_2$ respectively.

First, every ideal in $R$ is of the form $I×J$, where $I, J$ are ideals of $R_1, R_2$ respectively. In fact, if $\alpha$ is ideal of $R$, then $\alpha=(1,0)\alpha×(0,1)\alpha$.

Second, since $(1,0)(0,1)=(0,0)$, we know every prime ideal in $R$ contains $(1,0)$ or $(0,1)$.

So the result follows.