Is this proof on ideal maximality correct?

Question

The exercise is:

Given that $ \phi : R \to S $ is an onto ring homomorphism, let $ B $ be a maximal ideal of $ S $. Prove that $ A = \phi^{-1} (B) $ is a maximal ideal of $R$.

Ok, I previously proved that $A$ is an ideal of $R$, now to prove that it is maximal:

Let $I$ be an ideal of $R$ such that $ A \subset I$. From this follows that $B \subset \phi (I)$, and since $B$ is maximal $ \phi (I) = S$, and $I = R$, which means that $A$ is maximal.

Is it correct? Because I'm not too sure.

In general, $\phi(I)$ may not be an ideal in $S$. In your case, surjectivity of $\phi$ makes $\phi(I)$ an ideal. If you haven't proved it before, I would recommend proving this. Also, it's not clear to me why $\phi(I)=S\implies I=R$. — cqfd, Mar 05 '20 at 17:19
https://math.stackexchange.com/questions/1198414/pull-back-image-of-maximal-ideal-under-surjective-ring-homomorphism-is-maximal?rq=1 — cqfd, Mar 05 '20 at 17:21

egreg · Accepted Answer · 2020-03-06T16:17:37.503

3

You're a little too fast. Suppose $A\subset I$ (proper subset). Then $\phi(A)\subset\phi(I)$, which is an ideal because $\phi$ is surjective; since $\phi$ is surjective, $\phi(A)=B$. By maximality, either $\phi(I)=B$ or $\phi(I)=S$.

Now note that $\phi^{-1}(\phi(I))=I$, so we must have $I=R$.

Better technique: the homomorphism theorem provides an isomorphism $$ R/\phi^{-1}(B)\to S/B $$ The ideal $B$ is maximal if and only if $S/B$ is…

edited Mar 06 '20 at 16:17

answered Mar 05 '20 at 17:18

egreg

244,946

I used $\subset $ in the sense of "it is properly contained in", so $I \neq A$. – Dah_Fisicist Mar 06 '20 at 16:08
@Dah_Fisicist Fixed accordingly. – egreg Mar 06 '20 at 16:17

Is this proof on ideal maximality correct?

1 Answers1