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Let $R$ be a ring with identity. The ring $R$ is semisimple if it is semisimple as a left $R$ module. A module $M$ is semisimple if it can be expressed as a direct sum of simple submodules. The Jacobson radical of $R$, denoted by $J(R)$, is the intersection of all maximal left ideals of $R$.

I am asked to prove this:

The ring $R$ is semisimple if and only if it is Artinian and $J(R) = 0$.

I have proved the $\implies$ part. How to prove the converse? Since $R$ is Artinian every collection of left ideals of $R$ has a minimal element. How to go from here?

Jendrik Stelzner
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I'll make a try to write correctly an answer I came across:

Let $R$ Artinian with $rad(R)=0$ and $R$ not ss (semisimple). I choose $$I_1\subset R$$ minimal left ideal. It is $$I_1\not\subset 0=rad(R)$$ so $\exists$ $M_1$ maximal left ideal of $R$ such that $$I_1\not\subset M_1.$$ Therefore $I_1\cap M_1 \subset I_1 \Rightarrow I_1\cap M_1=0 $ and $M_1\subset I_1+M_1$ so $I_1\oplus M_1=R$. Since $M_1\not=0$ there exists a minimal left ideal $I_2\subset M_1$. Similarly I can find a maximal left ideal $N_2\subset R$ st $R=I_2\oplus N_2$. Since $I_2\subset M_1$ and $R=I_2\oplus N_2$ it is $M_1=I_2\oplus (N_2\cap M_1)$. So I write $M_2=N_2\cap M_1$ and I have $M_1=I_2\oplus M_2$.

So far we have $R=I_1\oplus I_2\oplus M_2$ and $M_2\not=0$. We repeat the same process. The sequence of left ideals $$R\supset M_1\supset M_2\supset ...$$ is strictly decreasing which is a contradiction.

Is this correct? Is there a shorter way to prove this?

IJM98
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