If the Localization is Noetherian for All Prime Ideals, Is the Ring Noetherian?
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?
Sponsored Links
Contents
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 $R$.
Add to solve later
Sponsored Links
More from my site
![The Ideal $(x)$ is Prime in the Polynomial Ring $R[x]$ if and only if the Ring $R$ is an Integral Domain](https://yutsumura.com/wp-content/uploads/2016/12/ring-theory-eye-catch-150x150.jpg)
The Ideal $(x)$ is Prime in the Polynomial Ring $R[x]$ if and only if the Ring $R$ is an Integral Domain
Let $R$ be a commutative ring with $1$. Prove that the principal ideal $(x)$ generated by the element $x$ in the polynomial ring $R[x]$ is a prime ideal if and only if $R$ is an integral domain.
Prove also that the ideal $(x)$ is a maximal ideal if and only if $R$ is a […]
Every Prime Ideal is Maximal if $a^n=a$ for any Element $a$ in the Commutative Ring
Let $R$ be a commutative ring with identity $1\neq 0$. Suppose that for each element $a\in R$, there exists an integer $n > 1$ depending on $a$.
Then prove that every prime ideal is a maximal ideal.
Hint.
Let $R$ be a commutative ring with $1$ and $I$ be an ideal […]- Every Maximal Ideal of a Commutative Ring is a Prime Ideal
Let $R$ be a commutative ring with unity.
Then show that every maximal ideal of $R$ is a prime ideal.
We give two proofs.
Proof 1.
The first proof uses the following facts.
Fact 1. An ideal $I$ of $R$ is a prime ideal if and only if $R/I$ is an integral […]
- If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field.
Let $R$ be a commutative ring with $1$.
Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.
Proof.
As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption.
Hence $R=R/\{0\}$ is an integral […]
- Every Prime Ideal in a PID is Maximal / A Quotient of a PID by a Prime Ideal is a PID
(a) Prove that every prime ideal of a Principal Ideal Domain (PID) is a maximal ideal.
(b) Prove that a quotient ring of a PID by a prime ideal is a PID.
Proof.
(a) Prove that every PID is a maximal ideal.
Let $R$ be a Principal Ideal Domain (PID) and let $P$ […]
- If $R$ is a Noetherian Ring and $f:R\to R’$ is a Surjective Homomorphism, then $R’$ is Noetherian
Suppose that $f:R\to R'$ is a surjective ring homomorphism.
Prove that if $R$ is a Noetherian ring, then so is $R'$.
Definition.
A ring $S$ is Noetherian if for every ascending chain of ideals of $S$
\[I_1 \subset I_2 \subset \cdots \subset I_k \subset […]
- Every Prime Ideal of a Finite Commutative Ring is Maximal
Let $R$ be a finite commutative ring with identity $1$. Prove that every prime ideal of $R$ is a maximal ideal of $R$.
Proof.
We give two proofs. The first proof uses a result of a previous problem. The second proof is self-contained.
Proof 1.
Let $I$ be a prime ideal […]
- Is the Quotient Ring of an Integral Domain still an Integral Domain?
Let $R$ be an integral domain and let $I$ be an ideal of $R$.
Is the quotient ring $R/I$ an integral domain?
Definition (Integral Domain).
Let $R$ be a commutative ring.
An element $a$ in $R$ is called a zero divisor if there exists $b\neq 0$ in $R$ such that […]
