If $R$ is an Artinian ring with $\{\mathfrak p_1,\ldots,\mathfrak p_n\}$ the set of prime and thus maximal ideals of $R$, is it true that $R_{\mathfrak p_i}$ (the localization at $\mathfrak p_i$) is isomorphic to $R/\mathfrak p_i^k$ for some $k\in\mathbb{N}_{\ge1}$ for all $i\in\{1,\ldots,n\}$?
I was trying to prove that $$\phi:R\to\prod_{\mathfrak p\in\text{Spec}(R)}R_{\mathfrak p}$$ is an isomorphism, and then the Chinese Remainder Theorem got the best off me and led me to conclude that $$R\cong\prod_{\mathfrak p\in\text{Spec}(R)}R/\mathfrak p^k,$$so if my assertion turns out to be true that would be really nice.
Does anyone have some insights on whether and why this is true or false?
Let $R$ be an artinian ring, and $\mathfrak m\in\operatorname{Max}R$. Then $R_{\mathfrak m}$ is local and artinian, so there is $k\ge1$ such that $(\mathfrak mR_{\mathfrak m})^k=\mathfrak m^kR_{\mathfrak m}=0$. Chose $k$ minimal with this property. (Note that $\mathfrak m^k=\mathfrak m^{k+1}=\cdots$.) Then the canonical homomorphism $f:R\to R_{\mathfrak m}$ gives rise to an isomorphism $R/\mathfrak m^k\simeq R_{\mathfrak m}$.
First of all let's show that $\ker f=\mathfrak m^k$. If $a\in R$ is such that $f(a)=\frac01$, then there is $s\in R\setminus\mathfrak m$ such that $sa=0$. From $sa\in\mathfrak m^k$ and $s\in R\setminus\mathfrak m$ we get $a\in\mathfrak m^k$. (Note that $\mathfrak m^k$ is $\mathfrak m$-primary.) Conversely, if $a\in\mathfrak m^k$ then obviously $f(a)=\frac 01$.
It remains to prove that $f$ is surjective. Let $s\in R\setminus\mathfrak m$. We want to show that there is $a\in R$ such that $f(a)=\frac 1s$. Chose $a\in R$ such that $sa-1\in\mathfrak m^k$, and then $\frac{sa-1}{1}=\frac01$ in $R_{\mathfrak m}$, so $\frac a1=\frac1s$. Since $s\in R\setminus\mathfrak m$ we get that $\bar s$ is invertible in $\bar R=R/\mathfrak m^k$ hence there is $\bar a\in\bar R$ such that $\bar s\bar a=\bar 1$, and we are done.