The Stacks project

Proof. Follows immediately from the more general Morphisms, Lemma 29.8.5. $\square$

Proof. This follows, upon taking affine opens, from Algebra, Lemma 10.30.2. (Of course it also follows from generic flatness.) $\square$

Proof. Let $Y' \subset Y$ be the reduction of $Y$. Set $X' = Y' \times _ Y X$ and $Z' = Y' \times _ Y Z$. As $Y' \to Y$ is a universal homeomorphism by Morphisms, Lemma 29.45.6 we see that it suffices to prove the lemma for $Z' \subset X' \to Y'$. Thus we may assume that $Y$ is integral, see Properties, Lemma 28.3.4. By Morphisms, Proposition 29.27.1 there exists a nonempty affine open $V \subset Y$ such that $X_ V \to V$ and $Z_ V \to V$ are flat and of finite presentation. We claim that $V$ works. Pick $y \in V$. If $Z_ y$ has a nonempty interior, then $Z_ y$ contains a generic point $\xi $ of an irreducible component of $X_ y$. Note that $\eta \leadsto f(\xi )$. Since $Z_ V \to V$ is flat we can choose a specialization $\xi ' \leadsto \xi $, $\xi ' \in Z$ with $f(\xi ') = \eta $, see Morphisms, Lemma 29.25.9. By Lemma 37.22.10 we see that

Hence some irreducible component of $Z_\eta $ passing through $\xi '$ has dimension $\dim _{\xi '}(X_\eta )$ which contradicts the assumption that $Z_\eta $ is nowhere dense in $X_\eta $ and we win. $\square$

Proof. Let $Y' \subset Y$ be the reduction of $Y$. Let $X' = Y' \times _ Y X$ and $U' = Y' \times _ Y U$. As $Y' \to Y$ induces a bijection on points, and as $U' \to U$ and $X' \to X$ induce isomorphisms of scheme theoretic fibres, we may replace $Y$ by $Y'$ and $X$ by $X'$. Thus we may assume that $Y$ is integral, see Properties, Lemma 28.3.4. We may also replace $Y$ by a nonempty affine open. In other words we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a domain with fraction field $K$.

As $f$ is of finite type we see that $X$ is quasi-compact. Write $X = X_1 \cup \ldots \cup X_ n$ for some affine opens $X_ i$. By Morphisms, Definition 29.7.1 we see that $U_ i = X_ i \cap U$ is an open subscheme of $X_ i$ such that $U_{i, \eta }$ is scheme theoretically dense in $X_{i, \eta }$. Thus it suffices to prove the result for the pairs $(X_ i, U_ i)$, in other words we may assume that $X$ is affine.

Write $X = \mathop{\mathrm{Spec}}(B)$. Note that $B_ K$ is Noetherian as it is a finite type $K$-algebra. Hence $U_\eta $ is quasi-compact. Thus we can find finitely many $g_1, \ldots , g_ m \in B$ such that $D(g_ j) \subset U$ and such that $U_\eta = D(g_1)_\eta \cup \ldots \cup D(g_ m)_\eta $. The fact that $U_\eta $ is scheme theoretically dense in $X_\eta $ means that $B_ K \to \bigoplus _ j (B_ K)_{g_ j}$ is injective, see Morphisms, Example 29.7.4. By Algebra, Lemma 10.24.4 this is equivalent to the injectivity of $B_ K \to \bigoplus \nolimits _{j = 1, \ldots , m} B_ K$, $b \mapsto (g_1b, \ldots , g_ mb)$. Let $M$ be the cokernel of this map over $A$, i.e., such that we have an exact sequence

After replacing $A$ by $A_ h$ for some nonzero $h$ we may assume that $B$ is a flat, finitely presented $A$-algebra, and that $M$ is flat over $A$, see Algebra, Lemma 10.118.3. The flatness of $B$ over $A$ implies that $B$ is torsion free as an $A$-module, see More on Algebra, Lemma 15.22.9. Hence $B \subset B_ K$. By assumption $I_ K = 0$ which implies that $I = 0$ (as $I \subset B \subset B_ K$ is a subset of $I_ K$). Hence now we have a short exact sequence

with $M$ flat over $A$. Hence for every homomorphism $A \to \kappa $ where $\kappa $ is a field, we obtain a short exact sequence

see Algebra, Lemma 10.39.12. Reversing the arguments above this means that $\bigcup D(g_ j \otimes 1)$ is scheme theoretically dense in $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa )$. As $\bigcup D(g_ j \otimes 1) = \bigcup D(g_ j)_\kappa \subset U_\kappa $ we obtain that $U_\kappa $ is scheme theoretically dense in $X_\kappa $ which is what we wanted to prove. $\square$

Proof. If $Y$ is Noetherian then $U = X \setminus (Z_1 \cup \ldots \cup Z_ n)$ is of finite type over $Y$ and we can directly apply Lemma 37.24.1 to get that $U_ V = \emptyset $ for a nonempty open $V \subset Y$. In general we argue as follows. As the question is topological we may replace $Y$ by its reduction. Thus $Y$ is integral, see Properties, Lemma 28.3.4. After shrinking $Y$ we may assume that $X \to Y$ is flat, see Morphisms, Proposition 29.27.1. In this case every point $x$ in $X_ y$ is a specialization of a point $x' \in X_\eta $ by Morphisms, Lemma 29.25.9. As the $Z_ i$ are closed in $X$ and cover the generic fibre this implies that $X_ y = \bigcup Z_{i, y}$ for $y \in Y$ as desired. $\square$

Proof. Choose an affine neighbourhood $\mathop{\mathrm{Spec}}(A) \subset Y$ of $\eta $. Choose an affine open $\mathop{\mathrm{Spec}}(B) \subset X$ mapping into $\mathop{\mathrm{Spec}}(A)$ via the morphism $f$. Let $\mathfrak p \subset A$ be the minimal prime corresponding to $\eta $. Let $B_{red}$ be the quotient of $B$ by the nilradical $\sqrt{(0)}$. The algebraic content of the lemma is that $C = B_{red} \otimes _ A \kappa (\mathfrak p)$ is reduced. Denote $I \subset A$ the nilradical so that $A_{red} = A/I$. Denote $\mathfrak p_{red} = \mathfrak p/I$ which is a minimal prime of $A_{red}$ with $\kappa (\mathfrak p) = \kappa (\mathfrak p_{red})$. Since $A \to B_{red}$ and $A \to \kappa (\mathfrak p)$ both factor through $A \to A_{red}$ we have $C = B_{red} \otimes _{A_{red}} \kappa (\mathfrak p_{red})$. Now $\kappa (\mathfrak p_{red}) = (A_{red})_{\mathfrak p_{red}}$ is a localization by Algebra, Lemma 10.25.1. Hence $C$ is a localization of $B_{red}$ (Algebra, Lemma 10.12.15) and hence reduced. $\square$

Proof. Let $V = \mathop{\mathrm{Spec}}(A)$ be a nonempty affine open of $Y$. By assumption the Jacobson radical of $A$ is a prime ideal $\mathfrak p$. Let $K = \kappa (\mathfrak p)$. Let $p$ be the characteristic of $K$ if positive and $1$ if the characteristic is zero. By Varieties, Lemma 33.6.11 there exists a finite purely inseparable field extension $K'/K$ such that $(X_{K'})_{red}$ is geometrically reduced over $K'$. Choose elements $x_1, \ldots , x_ n \in K'$ which generate $K'$ over $K$ and such that some $p$-power of $x_ i$ is in $A/\mathfrak p$. Let $A' \subset K'$ be the finite $A$-subalgebra of $K'$ generated by $x_1, \ldots , x_ n$. Note that $A'$ is a domain with fraction field $K'$. By Algebra, Lemma 10.46.7 we see that $A \to A'$ induces a universal homeomorphism on spectra. Set $Y' = \mathop{\mathrm{Spec}}(A')$. Set $X' = (Y' \times _ Y X)_{red}$. The generic fibre of $X' \to Y'$ is $(X_{K'})_{red}$ by Lemma 37.24.6 which is geometrically reduced by construction. Note that $X' \to X_ V$ is a finite universal homeomorphism as the composition of the reduction morphism $X' \to Y' \times _ Y X$ (see Morphisms, Lemma 29.45.6) and the base change of $g$. At this point all of the properties of the lemma hold except for possibly (7). This can be achieved by shrinking $Y'$ and hence $V$, see Morphisms, Proposition 29.27.1. $\square$

Proof. Let $V = \mathop{\mathrm{Spec}}(A)$ be a nonempty affine open of $Y$. By assumption the Jacobson radical of $A$ is a prime ideal $\mathfrak p$. Let $K = \kappa (\mathfrak p)$. By Varieties, Lemma 33.8.15 there exists a finite separable field extension $K'/K$ such that all irreducible components of $X_{K'}$ are geometrically irreducible over $K'$. Choose an element $\alpha \in K'$ which generates $K'$ over $K$, see Fields, Lemma 9.19.1. Let $P(T) \in K[T]$ be the minimal polynomial for $\alpha $ over $K$. After replacing $\alpha $ by $f \alpha $ for some $f \in A$, $f \not\in \mathfrak p$ we may assume that there exists a monic polynomial $T^ d + a_1T^{d - 1} + \ldots + a_ d \in A[T]$ which maps to $P(T) \in K[T]$ under the map $A[T] \to K[T]$. Set $A' = A[T]/(P)$. Then $A \to A'$ is a finite free ring map such that there exists a unique prime $\mathfrak q$ lying over $\mathfrak p$, such that $K = \kappa (\mathfrak p) \subset \kappa (\mathfrak q) = K'$ is finite separable, and such that $\mathfrak pA'_{\mathfrak q}$ is the maximal ideal of $A'_{\mathfrak q}$. Hence $g : Y' = \mathop{\mathrm{Spec}}(A') \to V = \mathop{\mathrm{Spec}}(A)$ is étale at $\mathfrak q$, see Algebra, Lemma 10.143.7. This means that there exists an open $W \subset \mathop{\mathrm{Spec}}(A')$ such that $g|_ W : W \to \mathop{\mathrm{Spec}}(A)$ is étale. Since $g$ is finite and since $\mathfrak q$ is the only point lying over $\mathfrak p$ we see that $Z = g(Y' \setminus W)$ is a closed subset of $V$ not containing $\mathfrak p$. Hence after replacing $V$ by a principal affine open of $V$ which does not meet $Z$ we obtain that $g$ is finite étale. $\square$