# Localizations of an Artinian ring are isomorphic to quotients.

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?

• Try to analyze the canonical homomorphism $R \rightarrow R_{p_i}$. – 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.

• @B.Pasternak If you like, I could give a one liner answer: $R/m^k$ is local with maximal ideal $m/m^k$, so $R/m^k\simeq(R/m^k)_{m/m^k}\simeq R_m/m^kR_m=R_m$. – 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 $$f(au)=f(a)f(u)=\frac a1 \frac 1b = \frac ab$$.