Lemma 10.20.3. Let $A \to B$ be a local homomorphism of local rings. Assume

1. $B$ is finite as an $A$-module,

2. $\mathfrak m_ B$ is a finitely generated ideal,

3. $A \to B$ induces an isomorphism on residue fields, and

4. $\mathfrak m_ A/\mathfrak m_ A^2 \to \mathfrak m_ B/\mathfrak m_ B^2$ is surjective.

Then $A \to B$ is surjective.

Proof. To show that $A \to B$ is surjective, we view it as a map of $A$-modules and apply Lemma 10.20.1 (6). We conclude it suffices to show that $A/\mathfrak m_ A \to B/\mathfrak m_ AB$ is surjective. As $A/\mathfrak m_ A = B/\mathfrak m_ B$ it suffices to show that $\mathfrak m_ AB \to \mathfrak m_ B$ is surjective. View $\mathfrak m_ AB \to \mathfrak m_ B$ as a map of $B$-modules and apply Lemma 10.20.1 (6). We conclude it suffices to see that $\mathfrak m_ AB/\mathfrak m_ A\mathfrak m_ B \to \mathfrak m_ B/\mathfrak m_ B^2$ is surjective. This follows from assumption (4). $\square$

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