# Prove that the preimage of a prime ideal is also prime.

Let $$f: R \rightarrow S$$ be a ring homomorphism, with $$R$$ and $$S$$ commutative and $$f(1)=1$$. If $$P$$ is a prime ideal of $$S$$, show that the preimage $$f^{-1}(P)$$ is a prime ideal of $$R$$.

Define $$g: S \rightarrow S/P$$ with kernel $$s$$. Let $$h = g \circ f: R \rightarrow S/P$$. Since $$h$$ is a ring homomorphism, the kernel is an ideal of $$R$$. Also, from the first isomorphism theorem, we know that $$R/\ker(h) \cong S/P$$. Since $$P$$ is a prime ideal of $$S$$, we know that $$S/P$$ is an integral domain. Since $$R/\ker(h)$$ is isomorphic to $$S/P$$, it must also be an integral domain, which implies that the kernel of $$h$$ (which is the preimage of $$P$$) is a prime ideal of $$R$$.

Do you think my answer is correct? The reason why I was a bit skeptical of my answer is because I did not use the fact that $$R$$ and $$S$$ are commutative. So I'm wondering if I missed something $$\dots$$

• You said "integral domain" which implies that at least $P/S$ is commutative. But you shouldn't worry about this, because the usual definition of a prime ideal assumes the whole ring is commutative to begin with, and this is sufficient to justify the presence of that hypothesis. Jun 3, 2013 at 9:57

I want to point out there's a far easier way of doing this: let $f\colon R\rightarrow S$ be a ring homomorphism and let $x,y\in R$ such that $xy\in f^{-1}(P)$. Then $f(x)f(y)=f(xy)\in P\implies f(x)\in P$ or $f(y)\in P$, since $P$ is prime, i.e. $x\in f^{-1}(P)$ or $y\in f^{-1}(P)$ and we are done.

• Great approach, but beware, It is also needed to prove that the preimage is not all the domain, since by definition prime ideals are not all the ring $R$. Jul 15, 2017 at 17:28
• @Santropedro Well, maybe. I mean, the fact that $f^{-1}(P)$ is not the whole $R$ is already implied by the fact that $P$ is a prime ideal in $S$. For, if $f^{-1}(P)=R$, then $f(1)\in P$. But, by definition of ring homomorphism, $f(1)=1$, so $1\in P$, against the fact that $P$ is a prime ideal in $S$. Jul 16, 2017 at 14:48
• You just clarified the doubt I had very concretly and perfectly, in fact I had the exact analogous doubt for preimage of maximal ideals, and your comment works exactly for that other doubt too! Thank you very much you helped me a ton! Jul 16, 2017 at 14:58
• @Santropedro You are very welcome :) Jul 17, 2017 at 15:00

What you've done is correct. The definition for prime ideals in commutative rings relies on commutativity.

For a non-commutative ring $R$, we have a different definition, and say that $P$ is a prime ideal if whenever the product of two ideals $IJ\subset P$, then either $I\subset P$ or $J\subset P$.

• Thanks a lot, but I think I missed the fact that R/kerf is isomorphic to a subring of S/P and not necessarily to S/P itself, right? As Tobias Kildetoft said here: math.stackexchange.com/questions/410266/… I'm just saying this, because I don't want others to be confused if they see my question.
– user58289
Jun 3, 2013 at 17:38

Sorry for digging up this old question, but there is a slight mistake in your proof. Your statement $R/\ker(h) \cong S/P$ is wrong; rather, $R/\ker(h) \cong h(R) \subseteq S/P$ where $h(R)$ is a subring of $S/P$. However the proof is saved by noticing that a subring of an integral domain is also an integral domain.

• This is clarified by the OP in a comment under the accepted answer. Jan 1, 2017 at 9:19
• Ah, yes! Should have looked more closely there. I'll keep this answer here just in case Jan 2, 2017 at 3:30