The Stacks project

Proof. Condition (4) implies $X$ is locally Noetherian (Morphisms, Lemma 29.15.6). The set of points at which morphism is quasi-finite is open (Morphisms, Lemma 29.56.2). A dense open of a locally Noetherian scheme contains all generic point of irreducible components, hence (4) implies (3). Condition (3) implies condition (1) by Morphisms, Lemma 29.20.5. Condition (1) implies condition (2). Thus it suffices to prove the lemma in case (2) holds.

Assume (2) holds. Recall that $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ is the set of points of $Y$ specializing to $y$, see Schemes, Lemma 26.13.2. Combined with Morphisms, Lemma 29.20.13 this shows we may replace $Y$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$. Thus we may assume $Y = \mathop{\mathrm{Spec}}(B)$ where $B$ is a Noetherian local ring of dimension $\leq 1$ and $y$ is the closed point.

Let $X = \bigcup X_ i$ be the irreducible components of $X$ viewed as reduced closed subschemes. If we can show each fibre $X_{i, y}$ is a discrete space, then $X_ y = \bigcup X_{i, y}$ is discrete as well and we conclude that $X \to Y$ is quasi-finite at all points of $X_ y$ by Morphisms, Lemma 29.20.6. Thus we may assume $X$ is an integral scheme.

If $X \to Y$ maps the generic point $\eta $ of $X$ to $y$, then $X$ is the spectrum of a finite extension of $\kappa (y)$ and the result is true. Assume that $X$ maps $\eta $ to a point corresponding to a minimal prime $\mathfrak q$ of $B$ different from $\mathfrak m_ B$. We obtain a factorization $X \to \mathop{\mathrm{Spec}}(B/\mathfrak q) \to \mathop{\mathrm{Spec}}(B)$. Let $x \in X$ be a point lying over $y$. By the dimension formula (Morphisms, Lemma 29.52.1) we have

We know that $\dim (B/\mathfrak q) = 1$, that the generic point of $X$ is not equal to $x$ and specializes to $x$ and that $R(X)$ is algebraic over $\kappa (\mathfrak q)$. Thus we get

Hence every point $x$ of $X_ y$ is closed in $X_ y$ by Morphisms, Lemma 29.20.2 and hence $X \to Y$ is quasi-finite at every point $x$ of $X_ y$ by Morphisms, Lemma 29.20.6 (which also implies that $X_ y$ is a discrete topological space). $\square$

Proof. By Lemma 33.17.1 the morphism $f$ is quasi-finite at every point of the fibre $X_ y$. Hence $X_ y$ is a discrete topological space (Morphisms, Lemma 29.20.6). As $f$ is proper the fibre $X_ y$ is quasi-compact, i.e., finite. Thus we can apply Cohomology of Schemes, Lemma 30.21.2 to conclude. $\square$

Proof. Part (1) follows from Morphisms, Lemma 29.51.6. Part (2) follows from part (3) and Lemma 33.17.2 (and the fact that finite morphisms have finite fibres).

Part (3) follows from part (4) and Morphisms, Lemma 29.20.7 but we will also give a direct proof. Let $x \in X$ be as in (3). By Cohomology of Schemes, Lemma 30.21.2 we may assume $f$ is finite. We may assume $X$ affine. This reduces us to the case of a finite birational morphism of Noetherian affine schemes $Y \to X$ and $x \in X$ such that $\mathcal{O}_{X, x}$ is a normal domain. Since $\mathcal{O}_{X, x}$ is a domain and $X$ is Noetherian, we may replace $X$ by an affine open of $x$ which is integral. Then, since $Y \to X$ is birational and $Y$ is reduced we see that $Y$ is integral. Writing $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$ we see that $A \subset B$ is a finite inclusion of domains having the same field of fractions. If $\mathfrak p \subset A$ is the prime corresponding to $x$, then $A_\mathfrak p$ being normal implies that $A_\mathfrak p \subset B_\mathfrak p$ is an equality. Since $B$ is a finite $A$-module, we see there exists an $a \in A$, $a \not\in \mathfrak p$ such that $A_ a \to B_ a$ is an isomorphism.

Let $x \in X$ and $y \in Y$ be as in (4). After replacing $X$ by an affine open neighbourhood we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $A \subset \mathcal{O}_{X, x}$, see Properties, Lemma 28.29.8. Then $A$ is a domain and hence $X$ is integral. Since $f$ is birational and $Y$ is reduced it follows that $Y$ is integral too. Consider the ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{Y, y}$. This is a ring map which is essentially of finite type, the residue field extension is finite, and $\dim (\mathcal{O}_{Y, y}/\mathfrak m_ x\mathcal{O}_{Y, y}) = 0$ (to see this trace through the definitions of quasi-finite maps in Morphisms, Definition 29.20.1 and Algebra, Definition 10.122.3). By Algebra, Lemma 10.124.2 $\mathcal{O}_{Y, y}$ is the localization of a finite $\mathcal{O}_{X, x}$-algebra $B$. Of course we may replace $B$ by the image of $B$ in $\mathcal{O}_{Y, y}$ and assume that $B$ is a domain with the same fraction field as $\mathcal{O}_{Y, y}$. Then $\mathcal{O}_{X, x} \subset B$ have the same fraction field as $f$ is birational. Since $\mathcal{O}_{X, x}$ is normal, we conclude that $\mathcal{O}_{X, x} = B$ (because finite implies integral), in particular, we see that $\mathcal{O}_{X, x} = \mathcal{O}_{Y, y}$. By Morphisms, Lemma 29.42.4 after shrinking $X$ we may assume there is a section $X \to Y$ of $f$ mapping $x$ to $y$ and inducing the given isomorphism on local rings. Since $X \to Y$ is closed (by Schemes, Lemma 26.21.11) necessarily maps the generic point of $X$ to the generic point of $Y$ it follows that the image of $X \to Y$ is $Y$. Then $Y = X$ and we've proved what we wanted to show. $\square$