Lemma 37.43.1. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is of finite type and separated. Let $S'$ be the normalization of $S$ in $X$, see Morphisms, Definition 29.53.3. Picture:

\[ \xymatrix{ X \ar[rd]_ f \ar[rr]_{f'} & & S' \ar[ld]^\nu \\ & S & } \]

Then there exists an open subscheme $U' \subset S'$ such that

$(f')^{-1}(U') \to U'$ is an isomorphism, and

$(f')^{-1}(U') \subset X$ is the set of points at which $f$ is quasi-finite.

**Proof.**
By Morphisms, Lemma 29.56.2 the subset $U \subset X$ of points where $f$ is quasi-finite is open. The lemma is equivalent to

$U' = f'(U) \subset S'$ is open,

$U = (f')^{-1}(U')$, and

$U \to U'$ is an isomorphism.

Let $x \in U$ be arbitrary. We claim there exists an open neighbourhood $f'(x) \in V \subset S'$ such that $(f')^{-1}V \to V$ is an isomorphism. We first prove the claim implies the lemma. Namely, then $(f')^{-1}V \cong V$ is both locally of finite type over $S$ (as an open subscheme of $X$) and for $v \in V$ the residue field extension $\kappa (v)/\kappa (\nu (v))$ is algebraic (as $V \subset S'$ and $S'$ is integral over $S$). Hence the fibres of $V \to S$ are discrete (Morphisms, Lemma 29.20.2) and $(f')^{-1}V \to S$ is locally quasi-finite (Morphisms, Lemma 29.20.8). This implies $(f')^{-1}V \subset U$ and $V \subset U'$. Since $x$ was arbitrary we see that (a), (b), and (c) are true.

Let $s = f(x)$. Let $(T, t) \to (S, s)$ be an elementary étale neighbourhood. Denote by a subscript ${}_ T$ the base change to $T$. Let $y = (x, t) \in X_ T$ be the unique point in the fibre $X_ t$ lying over $x$. Note that $U_ T \subset X_ T$ is the set of points where $f_ T$ is quasi-finite, see Morphisms, Lemma 29.20.13. Note that

\[ X_ T \xrightarrow {f'_ T} S'_ T \xrightarrow {\nu _ T} T \]

is the normalization of $T$ in $X_ T$, see Lemma 37.19.2. Suppose that the claim holds for $y \in U_ T \subset X_ T \to S'_ T \to T$, i.e., suppose that we can find an open neighbourhood $f'_ T(y) \in V' \subset S'_ T$ such that $(f'_ T)^{-1}V' \to V'$ is an isomorphism. The morphism $S'_ T \to S'$ is étale hence the image $V \subset S'$ of $V'$ is open. Observe that $f'(x) \in V$ as $f'_ T(y) \in V'$. Observe that

\[ \xymatrix{ (f'_ T)^{-1}V' \ar[r] \ar[d] & (f')^{-1}(V) \ar[d] \\ V' \ar[r] & V } \]

is a fibre square (as $S'_ T \times _{S'} X = X_ T$). Since the left vertical arrow is an isomorphism and $\{ V' \to V\} $ is a étale covering, we conclude that the right vertical arrow is an isomorphism by Descent, Lemma 35.23.17. In other words, the claim holds for $x \in U \subset X \to S' \to S$.

By the result of the previous paragraph we may replace $S$ by an elementary étale neighbourhood of $s = f(x)$ in order to prove the claim. Thus we may assume there is a decomposition

\[ X = V \amalg W \]

into open and closed subschemes where $V \to S$ is finite and $x \in V$, see Lemma 37.41.4. Since $X$ is a disjoint union of $V$ and $W$ over $S$ and since $V \to S$ is finite we see that the normalization of $S$ in $X$ is the morphism

\[ X = V \amalg W \longrightarrow V \amalg W' \longrightarrow S \]

where $W'$ is the normalization of $S$ in $W$, see Morphisms, Lemmas 29.53.10, 29.44.4, and 29.53.12. The claim follows and we win.
$\square$

## Comments (2)

Comment #7352 by Yijin Wang on

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