Lemma 33.17.1. Let $f : X \to Y$ be locally of finite type. Let $y \in Y$ be a point such that $\mathcal{O}_{Y, y}$ is Noetherian of dimension $\leq 1$. Assume in addition one of the following conditions is satisfied

1. for every generic point $\eta$ of an irreducible component of $X$ the field extension $\kappa (\eta )/\kappa (f(\eta ))$ is finite (or algebraic),

2. for every generic point $\eta$ of an irreducible component of $X$ such that $f(\eta ) \leadsto y$ the field extension $\kappa (\eta )/\kappa (f(\eta ))$ is finite (or algebraic),

3. $f$ is quasi-finite at every generic point of an irreducible component of $X$,

4. $Y$ is locally Noetherian and $f$ is quasi-finite at a dense set of points of $X$,

Then $f$ is quasi-finite at every point of $X$ lying over $y$.

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 29.55.2). 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

$\dim (\mathcal{O}_{X, x}) \leq \dim (B/\mathfrak q) + \text{trdeg}_{\kappa (\mathfrak q)}(R(X)) - \text{trdeg}_{\kappa (y)} \kappa (x)$

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

$1 \leq 1 - \text{trdeg}_{\kappa (y)} \kappa (x)$

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). $\square$

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