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?

  • $\begingroup$ Try to analyze the canonical homomorphism $R \rightarrow R_{p_i}$. $\endgroup$ – Manos Nov 16 '15 at 1:57

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.

  • 2
    $\begingroup$ @B.Pasternak If you like, I could give a one liner answer: $R/m^k$ is local with maximal ideal $m/m^k$, so $R/m^k\simeq(R/m^k)_{m/m^k}\simeq R_m/m^kR_m=R_m$. $\endgroup$ – user26857 Nov 16 '15 at 2:18

We prove that the localization map $f: R \rightarrow R_{\mathfrak p_1}$ gives rise to an isomorphism $R/\mathfrak p_1^k\simeq R_{\mathfrak p_1}$. First: what is $k$? It is well known that the nilradical of $R$ is nilpotent, $rad(0)^k = 0$ because $R$ is Artinian. $R/\mathfrak p_1^k$ is a local ring with unique maximal ideal $\mathfrak p_1/\mathfrak p_1^k$. Indeed, a maximal ideal of $\mathfrak p_1/\mathfrak p_1^k$ corresponds to a maximal ideal $\mathfrak m$ of $R$ containing $\mathfrak p_1^k$. Since $\mathfrak m$ is prime, this means that $\mathfrak m \supseteq \mathfrak p_1 $, hence $\mathfrak m = \mathfrak p_1 $. This means that the units in $R/\mathfrak p_1^k$ are the elements outside $\mathfrak p_1/\mathfrak p_1^k$.

Now, suppose $a \in ker f$, then $\frac a1=\frac 01$, hence $sa=0$ for an $s \in R\setminus \mathfrak p_1$. This means that $\bar s \bar a = \bar 0$ in $R/\mathfrak p_1^k$. But $\bar s \notin \mathfrak p_1/\mathfrak p_1^k$, therefore $\bar s$ is a unit in $ R/\mathfrak p_1^k$, thus $\bar a = \bar 0$, so $a \in \mathfrak p_1^k$.

Lemma If $b \in \mathfrak p_1^k $ then there is an $s \in R \setminus \mathfrak p_1$ with $sb=0$. Indeed, choose $b_j \in \mathfrak p_j \setminus \mathfrak p_1$ for $j= 2, \ldots , n$. Then $s=b_2^k\ldots b_n^k$ does the trick. It can't be in $\mathfrak p_1$ because $\mathfrak p_1$ is prime and $sb \in \mathfrak p_1^k \ldots \mathfrak p_n^k = (\mathfrak p_1 \cap \ldots \cap \mathfrak p_n)^k=rad(0)^k=0$.

It immediately follows from the lemma that $\mathfrak p_1^k \subseteq ker f$. So, we already have that $\mathfrak p_1^k = ker f$.

Now we show that $f$ is surjective. We want to find $x \in R $ such that $f(x)=\frac ab$ with $a \in R$ and $b \in R \setminus \mathfrak p_1$.

Suppose that the ideal in $R/\mathfrak p_1^k$ generated by $\bar b = b + \mathfrak p_1^k$ is proper, then it is contained in the maximal ideal $\mathfrak p_1 / \mathfrak p_1^k$ hence $b \in \mathfrak p_1$, a contradiction. Thus, $\bar 1 = \bar u \bar b$ from which follows that $ub-1 \in \mathfrak p_1^k$. The above lemma gives us $s \in R$ with $s(ub-1)=0$ hence $\frac u1 = \frac 1b$ in $R_{\mathfrak p_1}$. We have $f(au)=f(a)f(u)=\frac a1 \frac 1b = \frac ab$.


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