Preimage of Maximal Ideal of Finitely Generated Algebra is Maximal

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$


Also see