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I'm doing the preparation to an exam, and I'm stuck in the following:

If $R$ is a Noetherian ring with zero nilradical ($N(R) = 0$), and $S$ is the set of regular elements of $R$ ($r \in S$ if $rs = 0 \Rightarrow s = 0$) then $S^{-1}R$ is Artinian.

First, is easy to check that S is a multiplicative system, and the non-zero elements of $S^{-1}R$ are either units or zero-divisors. If one can show that the zero-divisors were nilpotents it will be over because that means that $S^{-1}R/N(R)= S^{-1}R $ is a field. But I don't know that is the case, since I can't prove this.

I tryed to show that if $P$ is prime ideal then $P$ should be maximal, by showing that $S^{-1}R/P$ is a field, but I can't do this too.

Any help will be welcome.

Thank you!

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  • $\begingroup$ possible duplicate of The total ring of fractions of a reduced Noetherian ring $\endgroup$ – user26857 Jan 20 '15 at 13:09
  • $\begingroup$ Dear @user26857, but being a product of fields do not implie that is artinian right? $\mathbb{R}\times\mathbb{R}$ is a domain that is a product of fields but $\mathbb{R}\times(O)$ is a ideal of height one. $\endgroup$ – User43029 Jan 20 '15 at 13:14
  • $\begingroup$ Ow sorry, $\mathbb{R}\times\mathbb{R}$ is not a domain. $\endgroup$ – User43029 Jan 20 '15 at 13:19
  • $\begingroup$ A ring is Artinian $\Leftrightarrow$ it is Noetherian and is of dimension zero. Now a finite product of fields is a Noetherian ring and also dimension zero. So it is Artinian. $\endgroup$ – Krish Jan 20 '15 at 13:21
  • $\begingroup$ @Charles I've posted a direct proof to your question. $\endgroup$ – user26857 Jan 20 '15 at 19:34
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$S^{-1}R$ is noetherian, reduced ($N(S^{-1}R)=0$), and every element of $S^{-1}R$ is a zerodivisor or invertible. Now your question follows from the following result:

Let $A$ be a reduced noetherian ring with the property that every element of $A$ is a zerodivisor or invertible. Then $A$ is artinian.

Let $\{\mathfrak p_1,\dots,\mathfrak p_n\}$ be the set of minimal primes of $A$. Since $A$ is reduced we have $(0)=\bigcap_{i=1}^n\mathfrak p_i$. Let $\mathfrak p$ be a prime ideal of $A$, and $x\in\mathfrak p$, $x\ne 0$. Then $x$ is a zerodivisor, hence there is $y\ne 0$ such that $xy=0$. In particular, $xy\in\mathfrak p_i$ for all $i=1,\dots,n$. If $x\notin\mathfrak p_i$ for all $i=1,\dots,n$, then $y\in\bigcap_{i=1}^n\mathfrak p_i$, so $y=0$, a contradiction. This shows that there is an $i$ such that $x\in \mathfrak p_i$. It follows that $\mathfrak p\subseteq\bigcup_{i=1}^n\mathfrak p_i$. By the prime avoidance lemma we get $\mathfrak p\subseteq\mathfrak p_i$ for some $i$, that is, $\mathfrak p=\mathfrak p_i$, and we are done.

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  • $\begingroup$ thank you, this helps a lot, I was having some issues with the other answer... $\endgroup$ – User43029 Jan 22 '15 at 23:06
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Here is another approach; it is more involved than user26857's, but may also help make things clearer (or maybe not; I've left several details to check):

The ring $R$ is reduced (i.e. has trivial nil-radical) and is Noetherian, so has finitely many minimal prime ideals, say $\mathfrak p_1, \dots, \mathfrak p_n$. Combining these facts, we find that $\mathfrak p_1 \cap \cdots \cap \mathfrak p_n = 0.$

From this, we find that the natural (diagonal) map $$R \hookrightarrow R/\mathfrak p_1 \times \cdots \times R/\mathfrak p_n$$ is an injection.

Now in general this won't be an isomorphism (consider the case $R = \mathbb C[x,y]/(xy)$).

However, after we invert $S$, it does become an isomorphism: i.e. we get an induced isomorphism $$ S^{-1} R \buildrel \cong \over \longrightarrow S^{-1} R/ \mathfrak p_1 (S^{-1} R) \times \cdots \times S^{-1} R/\mathfrak p_n (S^{-1} R) .$$

To check this, we use the fact that localizing is an exact functor. First of all, this exactness implies that after inverting $S$, we continue to have an injection. Secondly, if we consider the cokernel of the original injection, you can check that it is annihilated by an element of $S$, so that localizing it at $S$ gives zero, and so after localizing at $S$, our injection in fact becomes an isomorphism.

(Here is an element of $S$ which annihilates the cokernel: for each $i$, choose an element $x_i$ lying $\mathfrak p_i,$ but none of the $\mathfrak p_j$ for $j \neq i$ --- such an element exists by the same prime advoidance lemma used by user26857 --- and let $x = \sum_{i = 1}^n \prod_{j \neq i} x_j;$ then $x \in S$, and $x$ annihilates the cokernel of the original injection.)

Now one can check that each of the rings $S^{-1}R/\mathfrak p_i (S^{-1} R)$ is a field. One way to do this is as follows: first note that $S \cap \mathfrak p_i = \emptyset$, so that $S^{-1} R/\mathfrak p_i (S^{-1} \mathfrak R)$ embeds naturally into the fraction field of $R/\mathfrak p_i$. The key point is that this embedding is actually an isomorphism, so that $(S^{-1} R)/\mathfrak p_i (S^{-1} R)$ is naturally identified with the fraction field of $R/\mathfrak p_i$. To show this, you have to show that if $y \in R \setminus \mathfrak p_i,$ then you may find an element $y' \in S$ such that $y \equiv y' \bmod \mathfrak p_i$; the proof of this is similar to the construction of the element $x$ above.

In conclusion, $S^{-1} R$ is isomorphic to a finite product of fields; in fact, it is canonically isomorphic to the product of the fraction fields of the quotients $R/\mathfrak p$ as $\mathfrak p$ runs over the minimal prime ideals of $R$.


The structure of this argument becomes somewhat clearer if you think in geometric terms: the key point is that the Spec $R/\mathfrak p_i$ are the irreducible components of Spec $R/\mathfrak p$, and inverting $S$ is the same as localizing at the generic points of each of these components. But this more geometric way of thinking , while very natural and powerful, takes practice to learn.

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  • $\begingroup$ It's the same proof as the one given to the original question (see this answer), and this is exactly what I wanted to avoid (see my last comment under the question). $\endgroup$ – user26857 Jan 21 '15 at 6:27
  • $\begingroup$ @user26857: You're right, I hadn't looked at that answer. $\endgroup$ – tracing Jan 21 '15 at 12:16
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Since $R$ is Noetherian so is $S^{-1}R = Q(R)$. Therefore $Q(R)$ has a finite number of minimal primes $\{P_1,\ldots,P_n\}$ and $$ D = \{ \text{zerodivisors of } Q(R) \} = \bigcup_{i=1}^n P_i. $$ Furthermore in $Q(R)$ every element is either a unit or a zerodivisor, so every ideal $I$ of $Q(R)$ must be contained in the set of zerodivisors. By prime avoidance $I$ is contained in one of the minimal primes, hence every minimal prime is also maximal, or, equivalently, every prime ideal is maximal, or, equivalently, $\dim Q(R) = 0$. Now conclude with

$$ A \text{ is Artinian } \Longleftrightarrow A \text{ is Noetherian} + \dim A = 0 $$

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