The Stacks project

Proof. Observe that the set of points where $f$ is unramified is the same as the set of points where $f$ is étale and that this set is open. See Morphisms, Definitions 29.35.1 and 29.36.1 and Lemma 29.36.16. To check $f$ is étale at $x$ we may work étale locally on the base and on the target (Descent, Lemmas 35.23.29 and 35.31.1). Thus we can apply More on Morphisms, Lemma 37.41.1 and assume that $f : X \to Y$ is finite and that $x$ is the unique point of $X$ lying over $y = f(x)$. Then it follows that $f$ is finite locally free (Morphisms, Lemma 29.48.2).

Assume $f$ is finite locally free and that $x$ is the unique point of $X$ lying over $y = f(x)$. By Discriminants, Lemma 49.3.1 we find a locally principal closed subscheme $D_\pi \subset Y$ such that $y' \in D_\pi $ if and only if there exists an $x' \in X$ with $f(x') = y'$ and $f$ ramified at $x'$. Thus we have to prove that $y \not\in D_\pi $. Assume $y \in D_\pi $ to get a contradiction.

By condition (3) we have $\dim (\mathcal{O}_{X, x}) \geq 1$. We have $\dim (\mathcal{O}_{X, x}) = \dim (\mathcal{O}_{Y, y})$ by Algebra, Lemma 10.112.7. By Lemma 58.21.1 we can find $y' \in D_\pi $ specializing to $y$ with $\dim (\mathcal{O}_{Y, y'}) = 1$. Choose $x' \in X$ with $f(x') = y'$ where $f$ is ramified. Since $f$ is finite it is closed, and hence $x' \leadsto x$. We have $\dim (\mathcal{O}_{X, x'}) = \dim (\mathcal{O}_{Y, y'}) = 1$ as before. This contradicts property (4). $\square$