The Stacks project

Lemma 28.51.16. Let $f : Y \to X$ be a finite type morphism of schemes with $Y$ reduced and $X$ Nagata. Let $X'$ be the normalization of $X$ in $Y$. Let $x' \in X'$ be a point such that

  1. $\dim (\mathcal{O}_{X', x'}) = 1$, and

  2. the fibre of $Y \to X'$ over $x'$ is empty.

Then $\mathcal{O}_{X', x'}$ is a discrete valuation ring.

Proof. We can replace $X$ by an affine neighbourhood of the image of $x'$. Hence we may assume $X = \mathop{\mathrm{Spec}}(A)$ with $A$ Nagata. By Lemma 28.51.15 the morphism $X' \to X$ is finite. Hence we can write $X' = \mathop{\mathrm{Spec}}(A')$ for a finite $A$-algebra $A'$. By Lemma 28.51.7 after replacing $X$ by $X'$ we reduce to the case described in the next paragraph.

The case $X = X' = \mathop{\mathrm{Spec}}(A)$ with $A$ Noetherian. Let $\mathfrak p \subset A$ be the prime ideal corresponding to our point $x'$. Choose $g \in \mathfrak p$ not contained in any minimal prime of $A$ (use prime avoidance and the fact that $A$ has finitely many minimal primes, see Algebra, Lemmas 10.14.2 and 10.30.6). Set $Z = f^{-1}V(g) \subset Y$; it is a closed subscheme of $Y$. Then $f(Z)$ does not contain any generic point by choice of $g$ and does not contain $x'$ because $x'$ is not in the image of $f$. The closure of $f(Z)$ is the set of specializations of points of $f(Z)$ by Lemma 28.6.5. Thus the closure of $f(Z)$ does not contain $x'$ because the condition $\dim (\mathcal{O}_{X', x'}) = 1$ implies only the generic points of $X = X'$ specialize to $x'$. In other words, after replacing $X$ by an affine open neighbourhood of $x'$ we may assume that $f^{-1}V(g) = \emptyset $. Thus $g$ maps to an invertible global function on $Y$ and we obtain a factorization

\[ A \to A_ g \to \Gamma (Y, \mathcal{O}_ Y) \]

Since $X = X'$ this implies that $A$ is equal to the integral closure of $A$ in $A_ g$. By Algebra, Lemma 10.35.11 we conclude that $A_\mathfrak p$ is the integral closure of $A_\mathfrak p$ in $A_\mathfrak p[1/g]$. By our choice of $g$, since $\dim (A_\mathfrak p) = 1$ and since $A$ is reduced we see that $A_\mathfrak p[1/g]$ is a finite product of fields (the product of the residue fields of the minimal primes contained in $\mathfrak p$). Hence $A_\mathfrak p$ is normal (Algebra, Lemma 10.36.16) and the proof is complete. Some details omitted. $\square$


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