The Stacks project

Lemma 37.62.3 (Variant with separable extension). 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 and that for every generic point $\eta $ of an irreducible component of $X$ the field extension $\kappa (\eta )/K$ is separable. Then there exists a finite separable 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. This is proved in exactly the same manner as Theorem 37.62.1 with a few minor modifications. The most important change is to use More on Algebra, Lemma 15.116.9 instead of More on Algebra, Proposition 15.116.8. 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.

Since $X$ is flat over $A$ every generic point $\eta $ of $X$ maps to the generic point of $\mathop{\mathrm{Spec}}(A)$. After replacing $X$ by its reduction we may assume $X$ is reduced. In this case $X_ K$ is geometrically reduced over $K$ by Varieties, Lemma 33.6.8. Hence $X_ K \to \mathop{\mathrm{Spec}}(K)$ is smooth on a dense open by Varieties, Lemma 33.25.7. Thus the smooth locus $U \subset X$ of the morphism $X \to \mathop{\mathrm{Spec}}(A)$ is open (by Morphisms, Definition 29.34.1) and is dense in the generic fibre. This reduces us to the situation of the following 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 $\mathcal{O}_ i = \mathcal{O}_{X, x_ i}$. Observe that the fraction field of $\mathcal{O}_ i$ is the residue field of a generic point of $X$. Let $A_ i$ be the localization of $A$ at the maximal ideal corresponding to the image of $x_ i$ in $\mathop{\mathrm{Spec}}(A)$. We may apply More on Algebra, Lemma 15.116.9 and we find finite separable extensions $K_ i/K$ which are solutions for $A_ i \to \mathcal{O}_ i$. Let $L/K$ be a finite separable 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 (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. 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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