Pullbacks of maximal ideals in a finitely generated k-algebra are again maximal

A few days ago, some classmates and I were thinking about about whether or not morphisms in \mathrm{maxSpec}(A) were well defined. In particular, given a map of k-algebras f:A \to B, we consider the induced morphism f^*:\mathrm{maxSpec} A \to \mathrm{maxSpec B} given by X \mapsto f^{-1}(X). We all knew that pullbacks of maximal ideals need not be maximal for ring morphisms, which is precisely what led to a great deal of confusion– thankfully this was all fixed as soon as we were reminded that we were considering morphisms of k-algebras. While this was funny, I think the proof for this fact is both an elementary and interesting check up in commutative algebra:

Theorem 1: let A,B be finitely generated k-algebras. Then the pullback of any maximal ideal in a k-algebra homomorphism  is again maximal

We will nee two lemmas:

Lemma 1: Let A\subseteq B be a finite extension of rings with B a domain. Then A is a field if and only if B is a field.

Proof: Suppose that A is a field. Then B is a vector space over A and \mu_b: B \to B (multiplication by a non-zero element) is an A-linear endomorphism with trivial kernel (B is a domain!), and so it must be surjective as well. But then there exists some element a \in B so that ab=ba=1.

 

Now suppose B is a field. Choose a \in A to be nonzero, implying that \frac{1}{a} \in B. Then, by Cayley-Hamilton, there exist a_0\dots a_{d-1} \in A so that (\frac{1}{a})^d+ \dots+a_0=0. We clear denominators and see that \frac{1}{a} \in A.

 

Nice, so we have some good stuff. The next result is important and there are many proofs of it, so I won’t include any here, but here is the statement and a reference.

 

Lemma 2 (Zariski’s Lemma): if k is a field, and A:=k[x_1, \dots x_n] a finite integral domain over k and also a field, then A is an algebraic extension of k.

 

Proof: see here  for a one liner.

In particular, this means that for all maximal ideals m \in A, (with A a f.g k-algebra) we know that  A/m is a finite field extension of k. From this, the main theorem follows readily:

Proof of the main theorem:  let A,B be finitely generated k-algebras with B a domain and \phi:A \to B a k-algebra homomorphism. Then for each maximal ideal m, we have by Zariski’s lemma that B/m is a finite field extension of k. This in turn implies that A/\phi^{-1}(m) is also a finite field extension since A/ \phi^{-1}(m) \hookrightarrow B/m . But then by Lemma 1, it is a field, so \phi^{-1}(m) must have been maximal.

So, \mathrm{maxSpec} is a contravariant functor.

P.S.

Dig the title of this post. Now peek at the title of this blog. The little things matter :).