The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal

Problems and solutions of ring theory in abstract algebra

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

(1) $a+b\in J$,
(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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1 Response

  1. 08/11/2017

    […] Since the natural projections are surjective ring homomorphisms, the images $I$ and $J$ are ideals in $R$ and $S$, respectively. (see the post The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal.) […]

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