## Inverse image of a prime ideal is a prime ideal

Here is one of those problems that invariably shows up in homework for first course on ring theory: Assume $A$ and $B$ are commutative rings with 1. Prove that if $f: A\to B$ is a ring homomorphism, and $I$ is a prime ideal in $B$, then $f^{-1}(I)$ is a prime ideal in $A$.

Showing that $f^{-1}(I)$ is an ideal in $A$ is routine and I shall omit the details. To show that $f^{-1}(I)$ is a prime ideal in $A$, one way to proceed is to apply the definition of a prime ideal: if $xy\in f^{-1}(I)$, then we want to show that $x\in f^{-1}(I)$ and $y\in f^{-1}(I)$. I think this is the kind of approach that seems natural for a beginner in ring theory (like me). But recently, reading “Introduction to Commutative Algebra” by Atiyah & Macdonald, I learnt the following solution, which seems more insightful. The idea is as follows:

To show that $f^{-1}(I)$ is a prime ideal in $A$, it suffices to prove that $A/f^{-1}(I)$ is an integral domain. We claim that $A/f^{-1}(I)$ is isomorphic to a subring of $B/I$, and this would finish the proof immediately for the following reason: We know $B/I$ is an integral domain (because $I$ is a prime ideal in $B$) and every subring of an integral domain is again an integral domain. So we need to exhibit an injective ring homomorphism $\varphi: A/f^{-1}(I)\to B/I$. This map is explicitly given by $\displaystyle \varphi(a+f^{-1}(I))=f(a)+I$

For every $a, a'\in A$, we have $\varphi((a+f^{-1}(I))+(a'+f^{-1}(I)))=\varphi(a+a'+f^{-1}(I))=f(a+a')+I=f(a)+f(a')+I=(f(a)+I)+(f(a')+I)=\varphi(a+f^{-1}(I))+\varphi(a'+f^{-1}(I))$

and $\varphi((a+f^{-1}(I))(a'+f^{-1}(I)))=\varphi(aa'+f^{-1}(I))=f(aa')+I=f(a)f(a')+I=(f(a)+I)(f(a')+I)=\varphi(a+f^{-1}(I))\varphi(a'+f^{-1}(I))$

which shows that $f$ is a ring homomorphism. We will now show that $f$ is injective. Assume $\varphi(a)=\varphi(a')$ for some $a, a'\in A$. Then, by definition, $f(a)+I=f(a')+I$

or equivalently, $f(a-a')=f(a)-f(a')\in I$

which implies that $a-a'\in f^{-1}(I)$ so that $a+f^{-1}(I)=a'+f^{-1}(I)$. Thus, $\varphi$ is injective. We have proved that $\varphi: A/f^{-1}(I)\to B/I$ is an injective ring homomorphism. As a result, $A/f^{-1}(I)$ is isomorphic to a subring of $B/I$, and so is an integral domain, proving that $f^{-1}(I)$ is prime ideal in $A$.

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