# 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 $R’$.

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## 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.

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