Jacobson radical of an Artinian ring is nilpotent

Remarks

The chain $J \supseteq J^2 \supseteq J^3 \supseteq \cdots$ becomes constant, say $J^n = J^{n+1} = I$, and one shows $I = 0$. If $I \ne 0$, choose a left ideal $L$ minimal among those with $IL \ne 0$, and $x \in L$ with $Ix \ne 0$. Minimality gives $L = Rx$ and, since $I(Ix) = I^2 x = Ix \ne 0$, also $Ix = Rx$. So $x = yx$ for some $y \in I \subseteq J$; as $1 - y$ is a unit, $x = 0$, a contradiction. Example: for the ring of upper triangular $2 \times 2$ matrices over a field, $J$ is the set of strictly upper triangular matrices and $J^2 = 0$. The Artinian hypothesis is needed: the local ring $\Z_{(p)}$ is Noetherian, but $J = p\Z_{(p)}$ is not nilpotent, since $p^n \ne 0$ for all $n$.

Going deeper

For a student

This is a key step in the structure theory of Artinian rings. With the Artin–Wedderburn theorem, which describes $R/J$ as a product of matrix rings over division rings, it shows that every left Artinian ring is semiprimary. It is also used in the proof of the Hopkins–Levitzki theorem, which says that a left Artinian ring is left Noetherian.