If the Localization is Noetherian for All Prime Ideals, Is the Ring Noetherian?

Problems and solutions of ring theory in abstract algebra

Problem 617

Let $R$ be a commutative ring with $1$.
Suppose that the localization $R_{\mathfrak{p}}$ is a Noetherian ring for every prime ideal $\mathfrak{p}$ of $R$.
Is it true that $A$ is also a Noetherian ring?

 
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Proof.

The answer is no. We give a counterexample.
Let
\[R=\prod_{i=1}^{\infty}R_i,\] where $R_i=\Zmod{2}$.
As $R$ is not finitely generated, it is not Noetherian.


Note that every element $x\in R$ is idempotent, that is, we have $x^2=x$.

Let $\mathfrak{p}$ be a prime ideal in $R$.
Then $R/\mathfrak{p}$ is a domain and we have $x^2=x$ for any $x\in R/\mathfrak{p}$.
It follows that $x=0, 1$ and $R/\mathfrak{p} \cong \Zmod{2}$.
This also shows that every prime ideal in $R$ is maximal.


Now let us determine the localization $R_{\mathfrak{p}}$.

As the prime ideal $\mathfrak{p}$ does not contain any proper prime ideal (since every prime is maximal), the unique maximal ideal $\mathfrak{p}R_{\mathfrak{p}}$ of $R_{\mathfrak{p}}$ contains no proper prime ideals.

Recall that in general the intersection of all prime ideals is the ideal of all nilpotent elements.
Since $R_{\mathfrak{p}}$ does not have any nonzero nilpotent element, we see that $\mathfrak{p}R_{\mathfrak{p}}=0$.

(Remark: every Boolean ring has no nonzero nilpotent elements.)

It follows that $R_{\mathfrak{p}}$ is a filed, in particular an integral domain.

As before, since every element $x$ of $R_{\mathfrak{p}}$ satisfies $x^2=x$, we conclude that $x=0, 1$ and $R_{\mathfrak{p}} \cong \Zmod{2}$.


Since a field is Noetherian the localization $R_{\mathfrak{p}}$ is Noetherian for every prime ideal $\mathfrak{p}$ of $R$.

In summary, $R$ is not a Noetherian ring but the localization $R_{\mathfrak{p}}$ is Noetherian for every prime ideal $\mathfrak{p}$ of $R$.


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