A Noetherian ring is Artinian exactly when every prime ideal is maximal

If $R$ is Artinian, then Every prime ideal of an Artinian ring is maximal says that every prime ideal is maximal. The Noetherian hypothesis in the statement is then automatic from Every commutative Artinian ring is Noetherian.

Conversely, assume every prime ideal of $R$ is maximal. By A Noetherian ring has finitely many minimal prime ideals, the minimal primes over $(0)$ are ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r$ for some $r\ge 1$ unless $R=0$, in which case the conclusion is immediate. By hypothesis, each ${\mathfrak{p}}_i$ is maximal. Every prime ideal contains a minimal prime over $(0)$, so every prime ideal is one of the ${\mathfrak{p}}_i$. Hence The nilradical is the intersection of all prime ideals gives $N:=\mathrm{Nil}(R)={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_r$. Because distinct maximal ideals are comaximal, Chinese remainder theorem for pairwise comaximal ideals yields $R\mathrm{/}N\cong {\prod }_{i=1}^{r}R\mathrm{/}{\mathfrak{p}}_i$, and each factor is a field by $R\mathrm{/}M$ is a field if and only if $M$ is a maximal ideal.

By The nilradical of a Noetherian ring is nilpotent, choose $t\ge 1$ with $N^t=0$. Since $R$ is Noetherian, A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member makes each ideal $N^j$ finitely generated, so each quotient $N^j\mathrm{/}{N}^{j+1}$ is a finitely generated $R$-module. Because $N$ annihilates that quotient, it is a finitely generated $R\mathrm{/}N$-module. Under the product decomposition of step 1.2, this means a finite product of finitely generated vector spaces over the fields $R\mathrm{/}{\mathfrak{p}}_i$, so each quotient $N^j\mathrm{/}{N}^{j+1}$ has finite length. The same is true for $R\mathrm{/}N$ itself. Repeated use of Module length is additive in short exact sequences along $0\to N^{j+1}\to N^j\to N^j\mathrm{/}{N}^{j+1}\to 0$ therefore shows that $R$ has finite length as a module over itself.

By A commutative ring is Artinian exactly when it has finite length as a module over itself, a commutative ring has finite length as a module over itself exactly when it is Artinian. So step 2.1 proves that $R$ is Artinian. Together with step 1.1, this gives the asserted equivalence.