Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 85 / 4 / a / Solution

Use the descending chain condition on the finite intersections of maximal right ideals. It gives a minimal such intersection $K=M_1\cap \cdots \cap M_r$. Intersecting $K$ with any other maximal right ideal cannot decrease it, by minimality. Thus $K$ lies in every maximal right ideal, and $K=J(R)$. When $J(R)=0$, the diagonal mapis injective, and each summand on the right is a simple module.

For completeness, a submodule $N$ of a finite direct sum of simple modules is semisimple. Induct on the number of summands. For $S\oplus T$ with $S$ simple, $N\cap S$ is either $0$ or $S$. In the first case projection into $T$ identifies $N$ with a submodule of $T$, so induction applies. In the second case subtracting the $S$ coordinate shows $N=S\oplus (N\cap T)$, and induction again applies. This also shows finite module length. Apply it to the image of the diagonal map to obtainThis is the semisimplicity of a right Artinian ring with zero radical. It does not assume right Noetherianity in order to prove it.