**First proof of Lemma 37.27.5.**
We give two proofs of the lemma. These are essentially equivalent; the second is more self contained but a bit longer. Choose a diagram

\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X_ V \ar[r] \ar[d] & X \ar[d]^ f \\ Y' \ar[r]^ g & V \ar[r] & Y } \]

as in Lemma 37.24.7. Note that the generic fibre of $f'$ is the reduction of the generic fibre of $f$ (see Lemma 37.24.6) and hence is geometrically irreducible. Suppose that the lemma holds for the morphism $f'$. Then after shrinking $V$ all the fibres of $f'$ are geometrically irreducible. As $X' = (Y' \times _ V X_ V)_{red}$ this implies that all the fibres of $Y' \times _ V X_ V$ are geometrically irreducible. Hence by Lemma 37.27.2 all the fibres of $X_ V \to V$ are geometrically irreducible and we win. In this way we see that we may assume that the generic fibre is geometrically reduced as well as geometrically irreducible and we may assume $Y = \mathop{\mathrm{Spec}}(A)$ with $A$ a domain.

Let $x \in X_\eta $ be the generic point. As $X_\eta $ is geometrically irreducible and reduced we see that $L = \kappa (x)$ is a finitely generated extension of $K = \kappa (\eta )$ which is geometrically reduced and geometrically irreducible, see Varieties, Lemmas 33.6.2 and 33.8.6. In particular the field extension $L/K$ is separable, see Algebra, Lemma 10.44.2. Hence we can find $x_1, \ldots , x_{r + 1} \in L$ which generate $L$ over $K$ and such that $x_1, \ldots , x_ r$ is a transcendence basis for $L$ over $K$, see Algebra, Lemma 10.42.3. Let $P \in K(x_1, \ldots , x_ r)[T]$ be the minimal polynomial for $x_{r + 1}$. Clearing denominators we may assume that $P$ has coefficients in $A[x_1, \ldots , x_ r]$. Note that as $L$ is geometrically reduced and geometrically irreducible over $K$, the polynomial $P$ is irreducible in $\overline{K}[x_1, \ldots , x_ r, T]$ where $\overline{K}$ is the algebraic closure of $K$. Denote

\[ B' = A[x_1, \ldots , x_{r + 1}]/(P(x_{r + 1})) \]

and set $X' = \mathop{\mathrm{Spec}}(B')$. By construction the fraction field of $B'$ is isomorphic to $L = \kappa (x)$ as $K$-extensions. Hence there exists an open $U \subset X$, and open $U' \subset X'$ and a $Y$-isomorphism $U \to U'$, see Morphisms, Lemma 29.50.7. Here is a diagram:

\[ \xymatrix{ X \ar[rd] & U \ar[l] \ar@{=}[r] \ar[d] & U' \ar[r] \ar[d] & X' \ar[ld] \ar@{=}[r] & \mathop{\mathrm{Spec}}(B') \\ & Y \ar@{=}[r] & Y & } \]

Note that $U_\eta \subset X_\eta $ and $U'_\eta \subset X'_\eta $ are dense opens. Thus after shrinking $Y$ by applying Lemma 37.24.3 we obtain that $U_ y$ is dense in $X_ y$ and $U'_ y$ is dense in $X'_ y$ for all $y \in Y$. Thus it suffices to prove the lemma for $X' \to Y$ which is the content of Lemma 37.27.4.
$\square$

**Second proof of Lemma 37.27.5.**
Let $Y' \subset Y$ be the reduction of $Y$. Let $X' \to X$ be the reduction of $X$. Note that $X' \to X \to Y$ factors through $Y'$, see Schemes, Lemma 26.12.7. As $Y' \to Y$ and $X' \to X$ are universal homeomorphisms by Morphisms, Lemma 29.45.6 we see that it suffices to prove the lemma for $X' \to Y'$. Thus we may assume that $X$ and $Y$ are reduced. In particular $Y$ is integral, see Properties, Lemma 28.3.4. Thus by Morphisms, Proposition 29.27.1 there exists a nonempty affine open $V \subset Y$ such that $X_ V \to V$ is flat and of finite presentation. After replacing $Y$ by $V$ we may assume, in addition to (1), (2), (3) that $Y$ is integral affine, $X$ is reduced, and $f$ is flat and of finite presentation. In particular $f$ is universally open, see Morphisms, Lemma 29.25.10.

Pick a nonempty affine open $U \subset X$. Then $U \to Y$ is flat and of finite presentation with geometrically irreducible generic fibre. The complement $X_\eta \setminus U_\eta $ is nowhere dense. Thus after shrinking $Y$ we may assume $U_ y \subset X_ y$ is open dense for all $y \in Y$, see Lemma 37.24.3. Thus we may replace $X$ by $U$ and we reduce to the case where $Y$ is integral affine and $X$ is reduced affine, flat and of finite presentation over $Y$ with geometrically irreducible generic fibre $X_\eta $.

Write $X = \mathop{\mathrm{Spec}}(B)$ and $Y = \mathop{\mathrm{Spec}}(A)$. Then $A$ is a domain, $B$ is reduced, $A \to B$ is flat of finite presentation, and $B_ K$ is geometrically irreducible over the fraction field $K$ of $A$. In particular we see that $B_ K$ is a domain. Let $L$ be the fraction field of $B_ K$. Note that $L$ is a finitely generated field extension of $K$ as $B$ is an $A$-algebra of finite presentation. Let $K'/K$ be a finite purely inseparable extension such that $(L \otimes _ K K')_{red}$ is a separably generated field extension, see Algebra, Lemma 10.45.3. Choose $x_1, \ldots , x_ n \in K'$ which generate the field extension $K'$ over $K$, and such that $x_ i^{q_ i} \in A$ for some prime power $q_ i$ (proof existence $x_ i$ omitted). Let $A'$ be the $A$-subalgebra of $K'$ generated by $x_1, \ldots , x_ n$. Then $A'$ is a finite $A$-subalgebra $A' \subset K'$ whose fraction field is $K'$. Note that $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, see Algebra, Lemma 10.46.7. Hence it suffices to prove the result after base changing to $\mathop{\mathrm{Spec}}(A')$. We are going to replace $A$ by $A'$ and $B$ by $(B \otimes _ A A')_{red}$ to arrive at the situation where $L$ is a separably generated field extension of $K$. Of course it may happen that $(B \otimes _ A A')_{red}$ is no longer flat, or of finite presentation over $A'$, but this can be remedied by replacing $A'$ by $A'_ f$ for a suitable $f \in A'$, see Algebra, Lemma 10.118.3.

At this point we know that $A$ is a domain, $B$ is reduced, $A \to B$ is flat and of finite presentation, $B_ K$ is a domain whose fraction field $L$ is a separably generated field extension of the fraction field $K$ of $A$. By Algebra, Lemma 10.42.3 we may write $L = K(x_1, \ldots , x_{r + 1})$ where $x_1, \ldots , x_ r$ are algebraically independent over $K$, and $x_{r + 1}$ is separable over $K(x_1, \ldots , x_ r)$. After clearing denominators we may assume that the minimal polynomial $P \in K(x_1, \ldots , x_ r)[T]$ of $x_{r + 1}$ over $K(x_1, \ldots , x_ r)$ has coefficients in $A[x_1, \ldots , x_ r]$. Note that since $L/K$ is separable and since $L$ is geometrically irreducible over $K$, the polynomial $P$ is irreducible over the algebraic closure $\overline{K}$ of $K$. Denote

\[ B' = A[x_1, \ldots , x_{r + 1}]/(P(x_{r + 1})). \]

By construction the fraction fields of $B$ and $B'$ are isomorphic as $K$-extensions. Hence there exists an isomorphism of $A$-algebras $B_ h \cong B'_{h'}$ for suitable $h \in B$ and $h' \in B'$, see Morphisms, Lemma 29.50.7. In other words $X$ and $X' = \mathop{\mathrm{Spec}}(B')$ have a common affine open $U$. Here is a diagram:

\[ \xymatrix{ X = \mathop{\mathrm{Spec}}(B) \ar[rd] & U \ar[l] \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(B') = X' \ar[ld] \\ & Y = \mathop{\mathrm{Spec}}(A) & } \]

After shrinking $Y$ once more (by applying Lemma 37.24.3 to $Z = X \setminus U$ in $X$ and $Z' = X' \setminus U$ in $X'$) we see that $U_ y$ is dense in $X_ y$ and $U_ y$ is dense in $X'_ y$ for all $y \in Y$. Thus it suffices to prove the lemma for $X' \to Y$ which is the content of Lemma 37.27.4.
$\square$

## Comments (2)

Comment #4825 by Nicolas Müller on

Comment #5128 by Johan on