Let $R$ be a commutative ring with identity. Suppose that $R$ is semi-local with maximal ideals $P_1,...,P_n$. By the Chinese Remainder Theorem we have $$R/J(R) \simeq R/P_1 \times ... \times R/P_n$$ where $J(R)$ is the intersection of all the maximal ideals. Hence $R/J(R)$ is a direct sum of finite number of fields. Let $Soc(A)$ be the sum of all minimal ideals of a ring $A$. How can we show that $Soc(R/J(R))=R/J(R)$?.
On Wikipedia I found that $R$ is a semi-local ring exactly when $R/J(R)$ is semisimple, and that for any semisimple ring $R$ we have $Soc(R)=R$. Unfortunately, the details provided there are very sparse and as I am a beginner in Algebra, I am not familiar with semisimple rings. Can anyone shed some light on this matter?
Well, just to clarify the question, you are not asking about the socle of a semilocal ring. You are asking about the socle of $R/J$ where $R$ is a semilocal ring and $J$ is the Jacobson radical of $R$.
Semilocal rings do not even have to have a nonzero socle: an example would be any local domain that isn't a field. The socle can also be nonzero, or the entire ring. It just depends.
The classification of semisimple rings is a classical result in noncommutative ring theory, and you can find it in most books on the topic. To name a few, Isaacs Algebra: a graduate course, Lam's First course in noncommutative rings, Jacobson's Basic Algebra I, Basic Abstract Algebra by Bhattacharya and Jain, and for texts I am unfamiliar with and/or have forgotten.
The key result is called the Artin-Wedderburn theorem, and that is one of the things you'd aim to understand if you really wanted to understand semisimple rings.
There are lots of equivalent characterizations:
Now, if one just specializes to commutative rings, the Artin-Wedderburn theorem says "A commutative ring is semsimple if and only if it is a finite direct product of fields." In your case above, that is exactly what is happening to $R/J$: you said it was isomorphic to $\prod_{i=1}^n R/P_i$, where each of the $P_i$ were maximal ideals, and so each $R/P_i$ is a field.
But anyhow, to get back to the task at hand, it is not difficult to show that $R=Soc(R_R)$ when $R$ is a finite product of fields. (Let's just simplify the task you gave to this. There is no sense continuing to write $R/J$ everywhere.) Just write $R=\prod_{i=1}^n F_i$ for fields $F_i$.
First, let $I_i$ be the set of elements of $R$ which are zero everywhere except on the $i$th coordinate. You should show this is a minimal ideal of $R$. Then obviously $\sum_{i=1}^n I_i=R$, so $R\subseteq Soc(R_R)$. The reverse containment also holds, of course, so equality holds. So the only thing for you to do is to establish that $I_i$ is a minimal right ideal.