The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal
Problem 532
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 $S$.
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Proof.
As in the statement of the problem, let $I$ be an ideal of $R$.
Our goal is to show that the image $J=f(I)$ is an ideal of $S$.
For any $a,b\in J$ and $s\in S$, we need to show that
(2) $sa\in J$.
Since $a, b\in J=f(I)$, there exists $a’, b’\in I$ such that
\[f(a’)=a \text{ and } f(b’)=b.\]
Then we have
\begin{align*}
a+b=f(a’)+f(b’)=f(a’+b’)
\end{align*}
since $f$ is a homomorphism.
Since $I$ is an ideal, the sum $a’+b’$ is in $I$.
This yields that $a+b\in f(I)=J$, which proves (1).
Since $f:R\to S$ is surjective, there exists $r\in R$ such that $f(r)=s$.
It follows that
\[sa=f(r)f(a’)=f(ra’)\]
since $f$ is a homomorphism.
Since $I$ is an ideal of $R$, the product $ra’$ is in $I$.
Hence $sa\in f(I)=J$, and (2) is proved.
Therefore the image $J=f(I)$ is an ideal of $S$.
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