Every commutative Artinian ring is Noetherian

Let $N=\mathrm{Nil}(R)$. By An Artinian ring has only finitely many maximal ideals, the maximal ideals of $R$ are ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$ for some $r\ge 1$ unless $R=0$, in which case the conclusion is immediate. By The nilradical is the intersection of all prime ideals and Every prime ideal of an Artinian ring is maximal, one has $N={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_r$. Distinct maximal ideals are comaximal, so Chinese remainder theorem for pairwise comaximal ideals gives $R\mathrm{/}N\cong {\prod }_{i=1}^{r}R\mathrm{/}{\mathfrak{m}}_i$. Each factor is a field by $R\mathrm{/}M$ is a field if and only if $M$ is a maximal ideal, and a field is Noetherian because its only ideals are $0$ and itself. Repeated use of A product of two Noetherian rings is Noetherian therefore shows that $B:=R\mathrm{/}N$ is Noetherian.

By The nilradical of an Artinian ring is a nilpotent ideal, choose $t\ge 1$ with $N^t=0$. For each $0\le j<t$, put $M_j=N^j\mathrm{/}{N}^{j+1}$. Because $N{M}_j=0$, the action of $R$ on $M_j$ factors through $B=R\mathrm{/}N$, so $M_j$ is a $B$-module. Transport the standard idempotents of the product ring in step 1.1 to elements $e_1,\dots ,e_r\in B$. Then $M_j=e_1{M}_j\oplus \cdots \oplus e_r{M}_j$, and each summand is naturally a vector space over the field $R\mathrm{/}{\mathfrak{m}}_i$. If some $e_i{M}_j$ had no finite spanning set, recursively choose $v_1,v_2,\dots$ with $v_{n+1}\mathrm{\notin }\mathrm{span}(v_1,\dots ,v_n)$; then $\mathrm{span}(v_1,v_2,\dots )⊋\mathrm{span}(v_2,v_3,\dots )⊋\cdots$ would be a strict descending chain of $R$-submodules of $M_j$. But submodules of $M_j$ correspond to submodules of the ideal $N^j$ containing $N^{j+1}$, so this would give a strict descending chain of ideals in the Artinian ring $R$, impossible. Hence every $e_i{M}_j$ has a finite basis, and $M_j$ is finitely generated as a $B$-module.

Since $B$ is Noetherian, Finitely generated modules over a left Noetherian ring are Noetherian shows that each $M_j$ is Noetherian as a $B$-module, hence as an $R$-module. The filtration $0=N^t\subseteq N^{t-1}\subseteq \cdots \subseteq N\subseteq R$ has successive quotients $M_{t-1},\dots ,M_0$ and $R\mathrm{/}N=B$, all Noetherian. Repeatedly applying Noetherian and Artinian conditions are each exact in short exact sequences to the short exact sequences $0\to N^{j+1}\to N^j\to M_j\to 0$ shows that the regular module ${}_{R}R$ is Noetherian. Therefore Left and right Noetherian rings makes $R$ a Noetherian ring.