In an extension of finitely generated $k$-algebras the contraction of a maximal ideal is also maximal

Question

Let $k$ be a field and let $A \subset B$ be two finitely generated $k$-algebras. Prove that the contraction of any maximal ideal of $B$ is a maximal ideal of $A$.

thank you very much again!

Answer 1

Here is an outline of a proof for a more general fact. Let me know if you need more detail.

If $\varphi: A \to B$ is a ring map between $k$-algebras where $B$ is finitely generated, then the preimage of a maximal ideal in $B$ is a maximal ideal in $A.$

Proof:

1. For any ring map $\phi:R \to S$ and ideal $J\subseteq S$ there is a natural injective map $$\phi^*: \dfrac{R}{\phi^{-1}(J)} \to \dfrac{S}{J} \ : \ r + \phi^{-1}(J) \mapsto \phi(r) +J.$$
2. Zariski's Lemma: If $k\subseteq L$ is a field extension and $L=k[x_1,\ldots, x_n]$ for some $x_i\in L$ then $L$ is finite dimensional as a $k$ vector space. (i.e. For a field extension, "ring finite" $\implies$ "module finite").
3. There is an injection $k\to A/I$ for any proper ideal $I$ of $A.$
4. If $K\subseteq L$ is a finite field extension and $R$ is a ring such that $K\subseteq R \subseteq L$ then $R$ is a field as well.

Abuse notation a tiny little bit to regard our injections as subset inclusions, and for any maximal ideal $\mathfrak{m}\subset B$ we have the situation $$k\subseteq \frac{A}{\varphi^{-1}( \mathfrak{m})} \subseteq \frac{B}{\mathfrak{m}}.$$

Since $B$ is a finitely generated $k$ algebra, so is $\dfrac{B}{\mathfrak{m}}.$ By Zariski's lemma we have the situation of point 4 above, so $\varphi^{-1}(\mathfrak{m})$ is a maximal ideal.

Answer 2

An algebraic geometry way to see it is through Chevalley's theorem, which states that a finite type morphism of Noetherian schemes sends constructible sets to constructible sets. Let $\phi^{\#}: A \to B$ be an algebra homomorphism of finite type with $A$ and $B$ Noetherian. Then we have that $\phi: \operatorname{Spec}B \to \operatorname{Spec}A$ satisfies all the hypothesis of Chevalley's Theorem. We now note that the only constructible single element subsets of $\operatorname{Spec}C$ for any ring $C$ are those whose element is a maximal ideal. Thus if $m \in \operatorname{Spec}B$ is maximal we have that $\phi(\{m\}) = \{p\}$ for some $p \in \operatorname{Spec}A$. Since $\{p\}$ is constructible, $p$ is a maximal ideal. But we note that the definition of $\phi$ is $\phi(p) = {\phi^{\#}}^{-1}(p)$. We are done