The Stacks project

Proof. As the question is purely topological we may replace $X$ and $Y$ by their reductions. In particular this implies that $Y$ is integral, see Properties, Lemma 28.3.4. Let $X_\eta = X_{1, \eta } \cup \ldots \cup X_{n, \eta }$ be the decomposition of $X_\eta $ into irreducible components. Let $X_ i \subset X$ be the reduced closed subscheme whose generic fibre is $X_{i, \eta }$. Note that $Z_{i, j} = X_ i \cap X_ j$ is a closed subset of $X_ i$ whose generic fibre $Z_{i, j, \eta }$ is nowhere dense in $X_{i, \eta }$. Hence after shrinking $Y$ we may assume that $Z_{i, j, y}$ is nowhere dense in $X_{i, y}$ for every $y \in Y$, see Lemma 37.24.3. After shrinking $Y$ some more we may assume that $X_ y = \bigcup X_{i, y}$ for $y \in Y$, see Lemma 37.24.5. Moreover, after shrinking $Y$ we may assume that each $X_ i \to Y$ is flat and of finite presentation, see Morphisms, Proposition 29.27.1. The morphisms $X_ i \to Y$ are open, see Morphisms, Lemma 29.25.10. Thus there exists an open neighbourhood $V$ of $\eta $ which is contained in $f(X_ i)$ for each $i$. For each $y \in V$ the schemes $X_{i, y}$ are nonempty closed subsets of $X_ y$, we have $X_ y = \bigcup X_{i, y}$ and the intersections $Z_{i, j, y} = X_{i, y} \cap X_{j, y}$ are not dense in $X_{i, y}$. Clearly this implies that $X_ y$ has at least $n$ irreducible components. $\square$

Proof. This comes down to the statement that for $y' \in Y'$ with image $y \in Y$ the fibre $X'_{y'} = X_ y \times _ y y'$ is geometrically irreducible over $\kappa (y')$ if and only if $X_ y$ is geometrically irreducible over $\kappa (y)$. This follows from Varieties, Lemma 33.8.2. $\square$

Proof. Suppose that $y' \in Y'$ has image $y \in Y$. Suppose $K \supset \kappa (y)$ and $K' \supset \kappa (y')$ are separably closed extensions. Then we may choose a commutative diagram

of fields. The result follows as the morphisms of schemes

induce bijections between irreducible components, see Varieties, Lemma 33.8.7. $\square$

Proof. There exists an automorphism $\Psi $ of $A[x_1, \ldots , x_ n]$ over $A$ such that $\Psi (P) = ax_ n^ d +$ lower order terms in $x_ n$ with $a \not= 0$, see Algebra, Lemma 10.115.2. We may replace $P$ by $\Psi (P)$ and we may replace $A$ by $A_ a$. Thus we may assume that $P$ is monic in $x_ n$ of degree $d > 0$. For $i = 1, \ldots , n - 1$ let $d_ i$ be the degree of $P$ in $x_ i$. Note that this implies that $P^\varphi $ is monic of degree $d$ in $x_ n$ and has degree $\leq d_ i$ in $x_ i$ for every homomorphism $\varphi : A \to \kappa $ where $\kappa $ is a field. Thus if $P^\varphi $ is reducible, then we can write

with $Q_1, Q_2$ monic of degree $e_1, e_2 \geq 0$ in $x_ n$ with $e_1 + e_2 = d$ and having degree $\leq d_ i$ in $x_ i$ for $i = 1, \ldots , n - 1$. In other words we can write

where the sum is over the set $\mathcal{L}$ of multi-indices $L$ of the form $L = (l_1, \ldots , l_{n - 1})$ with $0 \leq l_ i \leq d_ i$. For any $e_1, e_2 \geq 0$ with $e_1 + e_2 = d$ we consider the $A$-algebra

where the $(\text{relations})$ is the ideal generated by the coefficients of the polynomial

with $Q_1$ and $Q_2$ defined as in (37.27.4.1). OK, and the assumption that $P$ is irreducible over $\overline{K}$ implies that there does not exist any $A$-algebra homomorphism $B_{e_1, e_2} \to \overline{K}$. By the Hilbert Nullstellensatz, see Algebra, Theorem 10.34.1 this means that $B_{e_1, e_2} \otimes _ A K = 0$. As $B_{e_1, e_2}$ is a finitely generated $A$-algebra this signifies that we can find an $f_{e_1, e_2} \in A$ such that $(B_{e_1, e_2})_{f_{e_1, e_2}} = 0$. By construction this means that if $\varphi : A_{f_{e_1, e_2}} \to \kappa $ is a homomorphism to a field, then $P^\varphi $ does not have a factorization $P^\varphi = Q_1 Q_2$ with $Q_1$ of degree $e_1$ in $x_ n$ and $Q_2$ of degree $e_2$ in $x_ n$. Thus taking $f = \prod _{e1, e_2 \geq 0, e_1 + e_2 = d} f_{e_1, e_2}$ we win. $\square$

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

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

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:

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

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:

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. $\square$

Proof. Let $Z$ be the reduced induced scheme structure on $\overline{\{ y\} }$. Let $f_ Z : X_ Z \to Z$ be the base change of $f$. Clearly it suffices to prove the lemma for $f_ Z$ and the generic point of $Z$. Hence we may assume that $Y$ is an integral scheme, see Properties, Lemma 28.3.4. Our goal in this case is to produce a nonempty open $V \subset Y$ such that $n_{X/Y}|_ V$ is constant.

We apply Lemma 37.24.8 to $f : X \to Y$ and we get $g : Y' \to V \subset Y$. As $g : Y' \to V$ is surjective finite étale, in particular open (see Morphisms, Lemma 29.36.13), it suffices to prove that there exists an open $V' \subset Y'$ such that $n_{X'/Y'}|_{V'}$ is constant, see Lemma 37.27.3. Thus we see that we may assume that all irreducible components of the generic fibre $X_\eta $ are geometrically irreducible over $\kappa (\eta )$.

At this point suppose that $X_\eta = X_{1, \eta } \bigcup \ldots \bigcup X_{n, \eta }$ is the decomposition of the generic fibre into (geometrically) irreducible components. In particular $n_{X/Y}(\eta ) = n$. Let $X_ i$ be the closure of $X_{i, \eta }$ in $X$. After shrinking $Y$ we may assume that $X = \bigcup X_ i$, see Lemma 37.24.5. After shrinking $Y$ some more we see that each fibre of $f$ has at least $n$ irreducible components, see Lemma 37.27.1. Hence $n_{X/Y}(y) \geq n$ for all $y \in Y$. After shrinking $Y$ some more we obtain that $X_{i, y}$ is geometrically irreducible for each $i$ and all $y \in Y$, see Lemma 37.27.5. Since $X_ y = \bigcup X_{i, y}$ this shows that $n_{X/Y}(y) \leq n$ and finishes the proof. $\square$

Proof. Fix $n$. Let $y \in Y$. We have to show that there exists an open neighbourhood $V$ of $y$ in $Y$ such that $E_ n \cap V$ is constructible in $V$. Thus we may assume that $Y$ is affine. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ of finite presentation whose base change to $Y$ recovers $f$. By Lemma 37.27.3 it suffices to prove the lemma for $f_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

We will use the criterion of Topology, Lemma 5.16.3 to prove that $E_ n$ is constructible in case $Y$ is a Noetherian scheme. To see this let $Z \subset Y$ be an irreducible closed subscheme. We have to show that $E_ n \cap Z$ either contains a nonempty open subset or is not dense in $Z$. Let $\xi \in Z$ be the generic point. Then Lemma 37.27.6 shows that $n_{X/Y}$ is constant in a neighbourhood of $\xi $ in $Z$. This clearly implies what we want. $\square$