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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 $\dots$

Thank you in advance.

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    $\begingroup$ 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. $\endgroup$
    – Marc van Leeuwen
    Jun 3, 2013 at 9:57

3 Answers 3

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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.

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    $\begingroup$ 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 $R$. $\endgroup$
    – Santropedro
    Jul 15, 2017 at 17:28
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    $\begingroup$ @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 $S$. $\endgroup$
    – Marco Vergura
    Jul 16, 2017 at 14:48
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    $\begingroup$ 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! $\endgroup$
    – Santropedro
    Jul 16, 2017 at 14:58
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    $\begingroup$ @Santropedro You are very welcome :) $\endgroup$
    – Marco Vergura
    Jul 17, 2017 at 15:00
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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 $J\subset P$.

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    $\begingroup$ 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. $\endgroup$
    – user58289
    Jun 3, 2013 at 17:38
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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.

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  • $\begingroup$ This is clarified by the OP in a comment under the accepted answer. $\endgroup$
    – user26857
    Jan 1, 2017 at 9:19
  • $\begingroup$ Ah, yes! Should have looked more closely there. I'll keep this answer here just in case $\endgroup$
    – JoeyBF
    Jan 2, 2017 at 3:30

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