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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