6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Try to analyze the canonical homomorphism $R \rightarrow R_{p_i}$. $\endgroup$ – Manos Nov 16 '15 at 1:57
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ @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
0
$\begingroup$

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 $f(au)=f(a)f(u)=\frac a1 \frac 1b = \frac ab$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.