An Artinian ring has only finitely many maximal ideals

If $R$ has no maximal ideals then the conclusion is immediate. Otherwise let $\mathcal{F}$ be the set of all finite nonempty intersections of maximal ideals of $R$. This set is nonempty, so [L1] gives a member minimal under inclusion; write it as $I={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_n$ for maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_n$.

Let $\mathfrak{m}$ be any maximal ideal of $R$. Then $I\cap \mathfrak{m}$ is again a finite nonempty intersection of maximal ideals, so $I\cap \mathfrak{m}\in \mathcal{F}$ and $I\cap \mathfrak{m}\subseteq I$. By the minimality from step 1.1, one has $I=I\cap \mathfrak{m}$, hence $I\subseteq \mathfrak{m}$. If $\mathfrak{m}$ were distinct from every ${\mathfrak{m}}_i$, then for each $1\le i\le n$ we could choose $x_i\in {\mathfrak{m}}_i\setminus \mathfrak{m}$ and put $x=x_1\cdots x_n$. Now $x\in I\subseteq \mathfrak{m}$, but every maximal ideal is prime by Every maximal ideal of a commutative ring is prime, so some factor $x_i$ would lie in $\mathfrak{m}$, contradicting the choice of the $x_i$. Therefore every maximal ideal of $R$ is one of ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_n$.

Therefore an Artinian ring has only finitely many maximal ideals.