Homomorphic image of principal ideal ring

Question

Question: show that homomorphic image of a principal ideal ring is a principal ideal ring.

My attempt : let $R$ be PIR (principal ideal ring) and $f$ be homomorphism from $R$ to a ring $S$ then by properties of ring homomorphism we know that $f(R)$ is subring of $S$. Now we consider an ideal $I$ of $f(R)$ then by properties of homomorphism we know its pullback that is $f^{-1}(I)$ is an ideal of $R$. But $R$ is PIR and hence there must exists some $a\in R$ such that $f^{-1}(I)=<a>$.

Hence to prove the result we just need to show that $I$ is principal ideal. My intension is that $I=<f(a)>$. But how to prove it?

Let $s\in <f(a)>$ then by definition of principal ideal, $s=s'f(a)$ for some $s'\in f(R)$. How to show $s\in I$? If $f(a)\in I$ then as $I$ is an ideal of $f(R)$ hence we have $s=s' f(a)\in I$ and we are done! But how $f(a)\in I$? I didnt get this! (Since $R$ my not have unity and hence $a$ does not belongs to $<a>=f^{-1}(I)$ and so we can't say $f(a)\in I$) and also how to show other direction that is, $I\subseteq <f(a)>$

Please help.

Answer 1

There seems to be a bit of ambiguity about what definitions you are using.

Since you make no mention of sides or commutativity, and you said $aR$ is an ideal, it looks like you are assuming commutativity.

$\langle a\rangle$ is, by definition (the standard definition), the smallest ideal of $R$ containing $a$, which would be $aR+a\mathbb Z$, in a commutative ring without identity.

Then from where you left off, I would continue that $f(aR+a\mathbb Z)=f(a)f(R)+f(a)\mathbb Z=\langle f(a)\rangle$ .

As far as I know, it is unheard of to call an ideal of the form $xR$ a principal ideal in a ring without identity. The problem is that you can never put your finger on a single element that generates $xR$ that way. (Pick any element $xr$ is $xrR=xR$? Who can tell?)

That's why we define $\langle a\rangle$ the other way so we can concretely name at least one element that generates it.

If you do in fact want to allow noncommutativity, you can use my same definition given above as the definition of a principle right ideal in a ring without identity. You'd have to make a corresponding one for principal left ideals. The arguments would still hold.

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Finally, in case the solution to the actual problem wasn't clear yet, the strategy you are using should be the right one. The single generator of the pre-image of the ideal will map to a single generator of the ideal, proving the image is a principal ideal ring. All of these arguments continue to work if you are working with one-sided principal ideal rings.

Answer 2

Definition: Let $R$ be a ring and $a\in R$. Then $$\langle a\rangle \text{ is smallest ideal containing }a \iff \langle a\rangle =\{ra+as+na+\sum_{i=1}^mr_ias_i\mid r,s,r_i,s_i\in R; n\in \Bbb{Z}; m\in \Bbb{N}\}.$$

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If $R$ is commutative and has an identity, then $\langle a\rangle =Ra=aR$. The definition you used for principle ideal assumes that $R$ is commutative with multiplicative identity. In general, $R$ may not have those additional structures.

If $f:R\to S$ is a homomorphism and $R$ is principal ideal ring, then $\text{Im}f$ is principal ideal ring.

Proof: It is easy to check $\text{Im}f$ is subring of $S$. Let $I$ be an ideal in $\text{Im}f$. Then $f^{-1}(I)$ is an ideal in $R$. Since $R$ is principal ideal ring, $f^{-1}(I)=\langle a\rangle$ for some $a\in R$. We claim $I=\langle f(a)\rangle$. Since $a\in \langle a\rangle =f^{-1}(I)$, we have $f(a)\in I$. Thus $\langle f(a)\rangle \subseteq I$. Let $y\in I\subseteq \text{Im}f$. Then $\exists x\in R$ such that $f(x)=y$. So $x\in f^{-1}(I)= \langle a\rangle$ and $x=ra+as+na+\sum_{i=1}^mr_ias_i$. Since $f$ is homomorphism, we have $$y=f(x)=f(r)f(a)+f(a)f(s)+nf(a)+\sum_{i=1}^mf(r_i)f(a)f(s_i)\in \langle f(a)\rangle.$$ Thus $I\subseteq \langle f(a)\rangle$ and $I=\langle f(a)\rangle$. Hence $\text{Im}f$ is principal ideal ring.