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 $\dots$

Thank you in advance.

Solution 1:

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.

Solution 2:

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 $J\subset P$.

Solution 3:

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.