Preimage of Maximal Ideal of Finitely Generated Algebra is Maximal

Theorem
Let $k$ be a field.

Let $A$ and $B$ be $k$-algebras.

Let $f : A \to B$ be a $k$-algebra homomorphism.

Let $B$ be finitely generated over $k$.

Let $\mathfrak m$ be a maximal ideal of $B$.

Then its preimage $f^{-1}(\mathfrak m)$ is a maximal ideal of $A$.

Proof
We have an injective morphism:


 * $\dfrac A {f^{-1} \left({\mathfrak m}\right)} \to \dfrac B {\mathfrak m}$

and thus $\dfrac A {f^{-1} \left({\mathfrak m}\right)}$ cannot have zero-divisors.

Thus $f^{-1} \left({\mathfrak m}\right)$ is prime.

We have that $\dfrac B {\mathfrak m}$ is a field extension of $k$ which is finitely generated

Thus, by Zariski's Lemma, $\dfrac B {\mathfrak m}$ is a finite field extension.

Hence the inverse of every element in $\dfrac B {\mathfrak m}$ can be written as a linear combination of the element.

Thus any sub-$k$-algebra must also be a field.

In particular $\dfrac A {f^{-1} \left({\mathfrak m}\right)}$ is a field.

Thus $f^{-1} \left({\mathfrak m}\right)$ is a maximal ideal.