The Stacks project

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 (37.65.0.1) 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. $\square$

Proof. Choose a finite affine open covering $S = \bigcup \mathop{\mathrm{Spec}}(A_ i)$. Then $K$ is equal to the fraction field of $A_ i$ for all $i$. Let $X_ i = X \times _ S \mathop{\mathrm{Spec}}(A_ i)$. Choose $L_ i/K$ as in Theorem 37.65.1 for the morphism $X_ i \to \mathop{\mathrm{Spec}}(A_ i)$. Let $B_ i \subset L_ i$ be the integral closure of $A_ i$ and let $Y_ i$ be the normalized base change of $X$ to $B_ i$. Let $L/K$ be a finite extension dominating each $L_ i$. Let $T_ i \subset T$ be the inverse image of $\mathop{\mathrm{Spec}}(A_ i)$. For each $i$ we get a commutative diagram

and in fact the left hand square is a normalized base change as discussed at the beginning of the section. In the proof of Theorem 37.65.1 we have seen that the smooth locus of $Y \to T$ contains the inverse image in $g^{-1}(T_ i)$ of the set of points where $Y_ i$ is smooth over $B_ i$. This proves the lemma. $\square$

Proof. This is proved in exactly the same manner as Theorem 37.65.1 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 37.24.3. 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 $\mathcal{O}_ i = \mathcal{O}_{X, x_ i}$. 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 (37.65.0.1) 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 (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. $\square$

Proof. This follows from Lemma 37.65.3 in exactly the same manner that Lemma 37.65.2 follows from Theorem 37.65.1. $\square$