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

Problems and solutions of ring theory in abstract algebra

Problem 413

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

 
FavoriteLoadingAdd to solve later

Sponsored Links

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.


FavoriteLoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Ring theory
Problems and solutions of ring theory in abstract algebra
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