Let $R$ be a semisimple artinian ring, i.e. a right artinian ring with no nonzero nilpotents right ideals.
We know that:
- $R\simeq M_{n_1}(D_1)\times\dots\times M_{n_t}(D_t)$ for suitable$t,n_1,\dots,n_t\in\Bbb Z_{\ge1}$, $D_j$'s division rings (Wedderburn Artin theorem).
- The right ideals of $R$ are exactly of the form $I_1\times\dots\times I_t$ where $I_j$ is a right ideal of $M_{n_j}(D_j)$.
I want to prove that homomorphic images of $R$ remains semisimple artinian rings.
So, let's consider a ring $S$ and a ring homomorphism $\varphi:R\to S$. We must check that $\varphi(R)$ is a semisimple artinian ring.
I have proved that $\varphi(R)$ is right artinian ring, showing that an arbitrary decreasing sequence of right ideals, becomes necessarely stationary.
My problem is to show that $\varphi(R)$ has no nonzero nilpotents right ideals! Here is my try: let's by contradiction consider $0\neq J\leq_{\mbox{r}}\varphi(R)$ be a nonzero nilpotent right ideal. Then $J=\varphi(I)$ for some $0\neq I\leq_{\mbox{r}}R$ (it's easy but not immediate).
Then $J^N=0,\;\exists N\ge1.$ Thus $0=J^N=\varphi(I^N)$, from which $I^N\le\ker\varphi$.
So I considered the descending chain $$ \ker\varphi\ge I^N\ge I^{N+1}\ge\cdots $$ in $R$ which is right artinian by hypotesis: thus there must exists $\overline N\ge1$ s.t. $I^{N+j}=I^{N+\overline N}\;\;\;\;\forall j\ge\overline N$. But $J\neq0\Rightarrow I\neq0$ so being $R$ semisimple artinian ring $I^{N+\overline N}\neq0$.
Here I stuck, I can't reach any contradiction.
Can someone help me? Many thanks!
If $I=\ker\varphi$, then $I$ is an ideal of $R=M_{n_1}(D_1)\times\dots\times M_{n_t}(D_t)$, so it is of the form $I=I_1\times\dots\times I_t$, where $I_j$ is an ideal of $M_{n_j}(D_j)$. Therefore $I_j=0$ or $I_j=M_{n_j}(D_j)$. Hence $\varphi(R)\cong R/I$ is a product of matrix rings over division rings, so it is semisimple artinian.