The preimage of a maximal ideal is maximal

I'm having some difficulty with this homework problem:

Let $A, B$ be reduced finitely generated $\mathbb{C}$-algebras, and $\psi : A \longrightarrow B$ a $\mathbb{C}$-algebra homomorphism. Let $\mathcal{M} \subset B$ be a maximal ideal. Show that $\psi^{-1}(\mathcal{M})$ is also a maximal ideal.

Here's what I have so far. The preimage of a prime ideal under a homomorphism is a prime ideal, so we know $\psi^{-1}(\mathcal{M})$ is a prime ideal. Also, if the homomorphism is surjective, then the image of an ideal is an ideal; but unfortunately we don't know $\psi$ is surjective.

So suppose that there is an ideal $\mathcal{J} \subset A$ such that $\mathcal{J} \supsetneq \psi^{-1}(\mathcal{M})$. Then $\psi(\mathcal{J}) \supsetneq \mathcal{M}$. But $\psi(\mathcal{J})$ is not necessarily an ideal, so consider the ideal it generates (which I'll denote $\langle\psi(\mathcal{J})\rangle$ - not sure if this notation is standard). This ideal contains $\mathcal{M}$, and since $\mathcal{M}$ is maximal we have $\langle\psi(\mathcal{J})\rangle = B$. Therefore any $f \in B$ can be written as a linear combination of elements of $B$ with coefficients in $\psi(\mathcal{J})$, and in particular, all of the generators of $B$ can be written this way.

I want to say that $\mathcal{J} = A$, but I'm not sure how to prove it.

• Following your idea, maybe you can prove that, if $S=Im\psi$ then $\mathcal{M}\cap S$ is a maximal ideal of $S$. If this is the case, then $\psi^{-1}(\mathcal{M})=\psi^{-1}(\mathcal{M}\cap S)$ and you may consider the co-restriction of $\psi$ to its image, which is surjective. I think this can't be done with general ring homomorphisms, but in this case you may use that $A$ and $B$ are $\mathbb{C}$-algebras. – Loronegro Feb 9 '12 at 3:08
• – Watson Nov 25 '18 at 18:57

First of all note that the assumption that $A$ and $B$ be algebras over a field $k$ and $\psi$ is a $k$-algebra homomorphism is crucial, since the injection $\mathbb Z \hookrightarrow \mathbb Q$ shows that the pullback of $(0)$, which is maximal in $\mathbb Q$, is $(0)$ which is not maximal in $\mathbb Z$. Or you can take $A=k[X] \hookrightarrow B=k(X)$, the natural injection of a polynomial ring into its function field and again pull back the zero ideal.
To prove this, since it's a homework problem, I'm going to assume the following lemma for a $k$-algebra $A$: (a) If $A$ is a domain such that every element $a \in A$ is the root of a non-zero polynomial in $k[X]$ (i.e. $A$ is algebraic over $k$), then $A$ is a field; (b) if $A$ is a field and is contained in an affine $k$-algebra domain, then $A$ is algebraic over $k$.
These are basic results which you should have covered, but I'll sketch the proof of this lemma before proceeding. For part (a), let $a \in A$, it suffices to show $k[a]$ is a field, and you may assume $A=k[a]$. Consider the map $k[X] \rightarrow A$ mapping $f$ to $f(a)$, let the kernel be $I$, show $A \cong k[X]/I$, and since $a$ is algebraic over $k$, $I \neq 0$. But $k[X]$ is a PID, so $I=(f)$ for $f$ irreducible, so $I$ is maximal and $A$ is a field. Part (b) takes more work, I'm afraid, but in the case $k=\mathbb C$ is algebraically closed, it's easier to see.
Given this lemma, define a map $A \rightarrow B/ \mathfrak m$ by sending $a$ to $\psi(a)+ \mathfrak m$, which has kernel $\psi^{-1}(\mathfrak m)=: \mathfrak n$. Hence, $A/ \mathfrak n$ is isomorphic to a subalgebra of $B/ \mathfrak m$. By part (b) of the lemma above, $B/ \mathfrak m$ is algebraic over $k$, hence $A/ \mathfrak n$ is also algebraic over $k$, and a domain, so by part (a) of the lemma, it's a field and so $\mathfrak n$ is maximal.