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

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.