# Preimage of Maximal Ideal of Finitely Generated Algebra is Maximal

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## 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:

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

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.

By Subalgebra of Finite Field Extension is Field, $\dfrac A {f^{-1} \left({\mathfrak m}\right)}$ is a field.

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

$\blacksquare$