Lemma 33.5.3. Notation as in Lemma 33.5.1. Assume $X$ is locally of finite type over $k$, that $\dim (\mathcal{O}_{X, x}) = \dim (\mathcal{O}_{Y, y})$ and that $\kappa (x) \otimes _ k K$ is reduced (for example if $\kappa (x)/k$ is separable or $K/k$ is separable). Then $\mathfrak m_ x \mathcal{O}_{Y, y} = \mathfrak m_ y$.

Proof. (The parenthetical statement follows from Algebra, Lemma 10.43.6.) Combining Lemmas 33.5.1 and 33.5.2 we see that $\mathcal{O}_{Y, y}/\mathfrak m_ x \mathcal{O}_{Y, y}$ has dimension $0$ and is reduced. Hence it is a field. $\square$

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