The Stacks project

Theorem 37.65.1. Let $A$ be a Dedekind ring with fraction field $K$. Let $X$ be a scheme flat and of finite type over $A$. Assume $A$ is a Nagata ring. There exists a finite extension $L/K$ such that the normalized base change $Y$ is smooth over $\mathop{\mathrm{Spec}}(B)$ at all generic points of all fibres.

Proof. During the proof we will repeatedly use that formation of the set of points where a (flat, finitely presented) morphism like $X \to \mathop{\mathrm{Spec}}(A)$ is smooth commutes with base change, see Morphisms, Lemma 29.34.15.

We first choose a finite extension $L/K$ such that $(X_ L)_{red}$ is geometrically reduced over $L$, see Varieties, Lemma 33.6.11. Since $Y \to (X_ B)_{red}$ is birational we see applying Varieties, Lemma 33.6.8 that $Y_ L$ is geometrically reduced over $L$ as well. Hence $Y_ L \to \mathop{\mathrm{Spec}}(L)$ is smooth on a dense open $V \subset Y_ L$ by Varieties, Lemma 33.25.7. Thus the smooth locus $U \subset Y$ of the morphism $Y \to \mathop{\mathrm{Spec}}(B)$ is open (by Morphisms, Definition 29.34.1) and is dense in the generic fibre. Replacing $A$ by $B$ and $X$ by $Y$ we reduce to the case treated in the next paragraph.

Assume $X$ is normal and the smooth locus $U \subset X$ of $X \to \mathop{\mathrm{Spec}}(A)$ is dense in the generic fibre. This implies that $U$ is dense in all but finitely many fibres, see Lemma 37.24.3. Let $x_1, \ldots , x_ r \in X \setminus U$ be the finitely many generic points of irreducible components of $X \setminus U$ which are moreover generic points of irreducible components of fibres of $X \to \mathop{\mathrm{Spec}}(A)$. Set $\mathcal{O}_ i = \mathcal{O}_{X, x_ i}$. Let $A_ i$ be the localization of $A$ at the maximal ideal corresponding to the image of $x_ i$ in $\mathop{\mathrm{Spec}}(A)$. By More on Algebra, Proposition 15.116.8 there exist finite extensions $K_ i/K$ which are solutions for the extension of discrete valuation rings $A_ i \to \mathcal{O}_ i$. Let $L/K$ be a finite extension dominating all of the extensions $K_ i/K$. Then $L/K$ is still a solution for $A_ i \to \mathcal{O}_ i$ by More on Algebra, Lemma 15.116.1.

Consider the diagram ( with the extension $L/K$ we just produced. Note that $U_ B \subset X_ B$ is smooth over $B$, hence normal (for example use Algebra, Lemma 10.163.9). Thus $Y \to X_ B$ is an isomorphism over $U_ B$. Let $y \in Y$ be a generic point of an irreducible component of a fibre of $Y \to \mathop{\mathrm{Spec}}(B)$ lying over the maximal ideal $\mathfrak m \subset B$. Assume that $y \not\in U_ B$. Then $y$ maps to one of the points $x_ i$. It follows that $\mathcal{O}_{Y, y}$ is a local ring of the integral closure of $\mathcal{O}_ i$ in $R(X) \otimes _ K L$ (details omitted). Hence because $L/K$ is a solution for $A_ i \to \mathcal{O}_ i$ we see that $B_\mathfrak m \to \mathcal{O}_{Y, y}$ is formally smooth in the $\mathfrak m_ y$-adic topology (this is the definition of being a "solution"). In other words, $\mathfrak m\mathcal{O}_{Y, y} = \mathfrak m_ y$ and the residue field extension is separable, see More on Algebra, Lemma 15.111.5. Hence the local ring of the fibre at $y$ is $\kappa (y)$. This implies the fibre is smooth over $\kappa (\mathfrak m)$ at $y$ for example by Algebra, Lemma 10.140.5. This finishes the proof. $\square$

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