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.
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