Proposition 1

Let A be a left Artinian ring. The radical $\mathfrak{r}$ of A is the largest nilpotent two-sided ideal of A, and the ring $A/\mathfrak{r}$ is semisimple.

We know (VIII, p. 156, Theorem 1) that every nilpotent two-sided ideal of A is contained in $\mathfrak{r}$. Let us prove that $\mathfrak{r}$ is nilpotent. Since the ring A is left Artinian and the sequence of two-sided ideals $\mathfrak{r}^n$ is decreasing, there exists an integer $p\geqslant 0$ such that we have $\mathfrak{r}^p=\mathfrak{r}^{p+1}$. Since A is Artinian, the A-module $A_s$ has finite length (VIII, p. 6, Theorem 1) and is therefore Noetherian. The left ideal $\mathfrak{r}^p$ is therefore finitely generated. By Nakayama’s lemma (Theorem 2 of VIII, p. 158), it follows that $\mathfrak{r}^p= 0$.

The ring $A/\mathfrak{r}$ is without radical (VIII, p. 155, Proposition 5); since it is left Artinian, it is semisimple (VIII, p. 154, Proposition 4).