Theorem 37.62.1. 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 37.23.3. 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 (37.62.0.1) 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: