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.