# If $R$ is a Noetherian Ring and $f:R\to R’$ is a Surjective Homomorphism, then $R’$ is Noetherian

## Problem 413

Suppose that $f:R\to R’$ is a surjective ring homomorphism.
Prove that if $R$ is a Noetherian ring, then so is $R’$.

Contents

## Definition.

A ring $S$ is Noetherian if for every ascending chain of ideals of $S$
$I_1 \subset I_2 \subset \cdots \subset I_k \subset \cdots$ there exists an integer $N$ such that we have
$I_N=I_{N+1}=I_{N+2}=\dots.$

## Proof.

To prove the ascending chain condition for $R’$, let
$I_1 \subset I_2 \subset \cdots \subset I_k \subset \cdots$ be an ascending chain of ideals of $R’$.
Note that the preimage $f^{-1}(I_k)$ of the ideal $I_k$ by a ring homomorphism is an ideal of $R$.
(See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a proof.)

Thus we obtain the ascending chain of ideals of $R$
$f^{-1}(I_1) \subset f^{-1}(I_2) \subset \cdots \subset f^{-1}(I_k) \subset \cdots.$ By assumption $R$ is Noetherian, and hence this ascending chain of ideals terminates. That is, there is an integer $N$ such that
$f^{-1}(I_N)=f^{-1}(I_{N+1})=f^{-1}(I_{N+2})=\dots.$

Since $f$ is surjective, we have
$f\left(\, f^{-1}(I_k) \,\right)=I_k$ for any $k$. Hence it follows that we have
$I_N=I_{N+1}=I_{N+2}=\dots.$ So each ascending chain of ideals of $R’$ terminates, and thus $R’$ is a Noetherian ring.

### More from my site

• The Preimage of Prime ideals are Prime Ideals Let $f: R\to R'$ be a ring homomorphism. Let $P$ be a prime ideal of the ring $R'$. Prove that the preimage $f^{-1}(P)$ is a prime ideal of $R$.   Proof. The preimage of an ideal by a ring homomorphism is an ideal. (See the post "The inverse image of an ideal by […]
• The Inverse Image of an Ideal by a Ring Homomorphism is an Ideal Let $f:R\to R'$ be a ring homomorphism. Let $I'$ be an ideal of $R'$ and let $I=f^{-1}(I)$ be the preimage of $I$ by $f$. Prove that $I$ is an ideal of the ring $R$.   Proof. To prove $I=f^{-1}(I')$ is an ideal of $R$, we need to check the following two […]
• A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring Let $R$ be the ring of all continuous functions on the interval $[0, 2]$. Let $I$ be the subset of $R$ defined by $I:=\{ f(x) \in R \mid f(1)=0\}.$ Then prove that $I$ is an ideal of the ring $R$. Moreover, show that $I$ is maximal and determine […]
• If the Localization is Noetherian for All Prime Ideals, Is the Ring Noetherian? 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?   Proof. The answer is no. We give a counterexample. Let […]
• The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal Let $R$ and $S$ be rings. Suppose that $f: R \to S$ is a surjective ring homomorphism. Prove that every image of an ideal of $R$ under $f$ is an ideal of $S$. Namely, prove that if $I$ is an ideal of $R$, then $J=f(I)$ is an ideal of $S$.   Proof. As in the […]
• Characteristic of an Integral Domain is 0 or a Prime Number Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.   Definition of the characteristic of a ring. The characteristic of a commutative ring $R$ with $1$ is defined as […]
• Generators of the Augmentation Ideal in a Group Ring Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by $\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,$ where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring […]
• $(x^3-y^2)$ is a Prime Ideal in the Ring $R[x, y]$, $R$ is an Integral Domain. Let $R$ be an integral domain. Then prove that the ideal $(x^3-y^2)$ is a prime ideal in the ring $R[x, y]$.   Proof. Consider the ring $R[t]$, where $t$ is a variable. Since $R$ is an integral domain, so is $R[t]$. Define the function $\Psi:R[x,y] \to R[t]$ sending […]

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### The Preimage of Prime ideals are Prime Ideals

Let $f: R\to R'$ be a ring homomorphism. Let $P$ be a prime ideal of the ring $R'$. Prove that...

Close