An Artinian ring is canonically the finite product of its localizations at its maximal ideals

If $R=0$, then the theorem has no further content. Assume from now on that $R\mathrm{\ne }0$. Let ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$ be the maximal ideals of $R$, finite by An Artinian ring has only finitely many maximal ideals, and put $N=\mathrm{Nil}(R)$. By The nilradical is the intersection of all prime ideals and Every prime ideal of an Artinian ring is maximal, one has $N={\mathfrak{m}}_1\cap \cdots \cap {\mathfrak{m}}_r$. Choose $n\ge 1$ with $N^n=0$ by The nilradical of an Artinian ring is a nilpotent ideal. Distinct maximal ideals are comaximal, and if $i\mathrm{\ne }j$ then choosing $a\in {\mathfrak{m}}_i$ and $b\in {\mathfrak{m}}_j$ with $a+b=1$ gives, from the binomial expansion of $(a+b{)}^{2n-1}$, the inclusion $1\in {\mathfrak{m}}_i^n+{\mathfrak{m}}_j^n$. So the ideals ${\mathfrak{m}}_i^n$ are pairwise comaximal. Also ${\prod }_{i=1}^r{\mathfrak{m}}_i^n={\left({\prod }_{i=1}^r{\mathfrak{m}}_i\right)}^n\subseteq N^n=0$. Hence Chinese remainder theorem for pairwise comaximal ideals yields $R\cong {\prod }_{i=1}^{r}R\mathrm{/}{\mathfrak{m}}_i^n$.

For each $i$, set $A_i=R\mathrm{/}{\mathfrak{m}}_i^n$. This quotient is local. Indeed Prime ideals of a quotient ring are exactly the prime ideals containing the ideal identifies prime ideals of $A_i$ with prime ideals $\mathfrak{p}$ of $R$ containing ${\mathfrak{m}}_i^n$. Since $N^n=0\subseteq {\mathfrak{m}}_i^n\subseteq \mathfrak{p}$, the prime $\mathfrak{p}$ contains $N$ and therefore equals one of the maximal ideals ${\mathfrak{m}}_j$. If $j\mathrm{\ne }i$, choose $a\in {\mathfrak{m}}_i$ and $b\in {\mathfrak{m}}_j$ with $a+b=1$; the same binomial argument as in step 1.1 gives $1\in {\mathfrak{m}}_i^n+{\mathfrak{m}}_j$. So ${\mathfrak{m}}_i^n⊈{\mathfrak{m}}_j$, a contradiction. Thus $\mathfrak{p}={\mathfrak{m}}_i$, and $A_i$ has the unique maximal ideal ${\mathfrak{m}}_i\mathrm{/}{\mathfrak{m}}_i^n$.

Let ${\lambda }_i:R\to R_{{\mathfrak{m}}_i}$ be localization and $q_i:R\to A_i$ the quotient map. Since $A_i$ is local with maximal ideal ${\mathfrak{m}}_i\mathrm{/}{\mathfrak{m}}_i^n$, every $s\mathrm{\notin }{\mathfrak{m}}_i$ maps to a unit of $A_i$, so Universal property of localisation: maps that invert $S$ factor uniquely through $S^{-1}R$ gives a unique homomorphism ${\psi }_i:R_{{\mathfrak{m}}_i}\to A_i$ with $q_i={\psi }_i{\lambda }_i$. Conversely, for each $j\mathrm{\ne }i$ choose $u_j\in {\mathfrak{m}}_j\setminus {\mathfrak{m}}_i$ and put $u={\prod }_{j\mathrm{\ne }i}{u}_j^n$. Then $u$ becomes a unit in $R_{{\mathfrak{m}}_i}$, while $u{\mathfrak{m}}_i^n\subseteq {\prod }_{j=1}^r{\mathfrak{m}}_j^n=0$ by step 1.1. Hence ${\lambda }_i$ kills ${\mathfrak{m}}_i^n$, and A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring gives a unique homomorphism ${\bar{\lambda }}_i:A_i\to R_{{\mathfrak{m}}_i}$ with ${\lambda }_i={\bar{\lambda }}_i{q}_i$. Both composites ${\psi }_i{\bar{\lambda }}_i$ and ${\bar{\lambda }}_i{\psi }_i$ agree with the identity after precomposing with the relevant universal map, so uniqueness in the same two universal properties forces them to be identities. Therefore $A_i\cong R_{{\mathfrak{m}}_i}$.

Combining steps 1.1 and 3.1 gives $R\cong {\prod }_{i=1}^r{R}_{{\mathfrak{m}}_i}$ canonically, and also $R\cong {\prod }_{i=1}^{r}R\mathrm{/}{\mathfrak{m}}_i^n$. The second form depends on the chosen nilpotence exponent $n$, while the first is the canonical product of localizations.