The Stacks project

Proof. It is clear that (1) implies (2) and (2) implies (3). Let us prove that (3) implies (1). Suppose that the base change $X_ T \to T$ is not open for some morphism of schemes $g : T \to S$. Then we can find some affine opens $V \subset S$, $U \subset X$, $W \subset T$ with $f(U) \subset V$ and $g(W) \subset V$ such that $U \times _ V W \to W$ is not open. If we can show that this implies $\mathbf{A}^ n \times U \to \mathbf{A}^ n \times V$ is not open, then $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is not open and the proof is complete. This reduces us to the result proved in the next paragraph.

Let $A \to B$ be a ring map such that $A' \to B' = A' \otimes _ A B$ does not induce an open map of spectra for some $A$-algebra $A'$. As the principal opens give a basis for the topology of $\mathop{\mathrm{Spec}}(B')$ we conclude that the image of $D(g)$ in $\mathop{\mathrm{Spec}}(A')$ is not open for some $g \in B'$. Write $g = \sum _{i = 1, \ldots , n} a'_ i \otimes b_ i$ for some $n$, $a'_ i \in A'$, and $b_ i \in B$. Consider the element $h = \sum _{i = 1, \ldots , n} x_ i b_ i$ in $B[x_1, \ldots , x_ n]$. Assume that $D(h)$ maps to an open subset under the morphism

in order to get a contradiction. Then $D(h)$ would map surjectively onto a quasi-compact open $U \subset \mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ n])$. Let $A[x_1, \ldots , x_ n] \to A'$ be the $A$-algebra homomorphism sending $x_ i$ to $a'_ i$. This also induces a $B$-algebra homomorphism $B[x_1, \ldots , x_ n] \to B'$ sending $h$ to $g$. Since

is cartesian the image of $D(g)$ in $\mathop{\mathrm{Spec}}(A')$ is equal to the inverse image of $U$ in $\mathop{\mathrm{Spec}}(A')$ and hence open which is the desired contradiction. $\square$

Proof. For any $n$ the scheme $\mathbf{A}^ n \times Y$ is geometrically unibranch by Lemma 37.36.4 and Properties, Lemma 28.15.6. Hence the hypotheses of the lemma hold for the morphisms $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times Y$ for all $n$. By Lemma 37.74.1 it suffices to prove $f$ is open. By Morphisms, Lemma 29.23.2 it suffices to show that generalizations lift along $f$. Suppose that $y' \leadsto y$ is a specialization of points in $Y$ and $x \in X$ is a point mapping to $y$. As in Lemma 37.41.1 choose a diagram

where $(V, v) \to (Y, y)$ is an elementary étale neighbourhood, $U \to V$ is finite, $u$ is the unique point of $U$ mapping to $v$, $U \subset V \times _ Y X$ is open, and $v \mapsto y$ and $u \mapsto x$. Let $E$ be an irreducible component of $U$ passing through $u$ (there is at least one of these). Since $U \to X$ is étale, $E$ maps to an irreducible component of $X$, which in turn dominates an irreducible component of $Y$ (by assumption). Since $U \to V$ is finite hence closed, we conclude that the image $E' \subset V$ of $E$ is an irreducible closed subset passing through $v$ which dominates an irreducible component of $Y$. Since $V \to Y$ is étale $E'$ must be an irreducible component of $V$ passing through $v$. Since $Y$ is geometrically unibranch we see that $E'$ is the unique irreducible component of $V$ passing through $v$ (Lemma 37.36.2). Since $V$ is locally Noetherian we may after shrinking $V$ assume that $E' = V$ (equality of sets).

Since $V \to Y$ is étale we can find a specialization $v' \leadsto v$ whose image is $y' \leadsto y$. By the above we can find $u' \in U$ mapping to $v'$. Then $u' \leadsto u$ because $u$ is the only point of $U$ mapping to $v$ and $U \to V$ is closed. Then finally the image $x' \in X$ of $u'$ is a point specializing to $x$ and mapping to $y'$ and the proof is complete. $\square$

Proof. If $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is universally open, then of course the image of $D(h)$ is open. Conversely, assume the image $U$ of $D(h)$ is open. Let $A \to A'$ be a ring map. It suffices to show that the image of any principal open $D(g) \subset \mathop{\mathrm{Spec}}(A' \otimes _ A B)$ in $\mathop{\mathrm{Spec}}(A')$ is open. We may write $g = \sum _{i = 1, \ldots , d} a'_ i \otimes b_ i$ for some $a'_ i \in A'$. Let $A[x_1, \ldots , x_ n] \to A'$ be the $A$-algebra homomorphism sending $x_ i$ to $a'_ i$. This also induces a $B$-algebra homomorphism $B[x_1, \ldots , x_ n] \to A' \otimes _ A B$ sending $h$ to $g$. Since

is cartesian the image of $D(g)$ in $\mathop{\mathrm{Spec}}(A')$ is equal to the inverse image of $U$ in $\mathop{\mathrm{Spec}}(A')$ and hence open. $\square$

Proof. By Limits, Lemma 32.18.2 after increasing $0$ we may assume $f_0$ is locally quasi-finite. Let $x \in X$. By étale localization of quasi-finite morphisms we can find a diagram

where $V \to Y$ is étale, $U \subset X_ V$ is open, $U \to V$ is finite, and $x$ is in the image of $U \to X$, see Lemma 37.41.1. After shrinking $V$ we may assume $V$ and $U$ are affine. Since $X$ is quasi-compact, it follows, by taking a finite disjoint union of such $V$ and $U$, that we can make a diagram as above such that $U \to X$ is surjective. By Limits, Lemmas 32.10.1, 32.4.11, 32.8.15, 32.8.3, 32.8.10, and 32.4.13 after possibly increasing $0$ we may assume we have a diagram

where $V_0$ is affine, $V_0 \to Y_0$ is étale, $U_0 \subset (X_0)_{V_0}$ is open, $U_0 \to V_0$ is finite, and $U_0 \to X_0$ is surjective. Since $V_ i \to Y_ i$ is étale and hence universally open, follows that it suffices to prove that $U_ i \to V_ i$ is universally open for large enough $i$. This reduces us to the case discussed in the next paragraph.

Let $A = \mathop{\mathrm{colim}}\nolimits A_ i$ be a filtered colimit of rings. Let $A_0 \to B_0$ be a ring map. Set $B = A \otimes _{A_0} B_0$ and $B_ i = A_ i \otimes _{A_0} B_0$. Assume $A_0 \to B_0$ is finite, of finite presentation, and $A \to B$ is universally open. We have to show that $A_ i \to B_ i$ is universally open for $i$ large enough. Pick $b_{0, 1}, \ldots , b_{0, d} \in B_0$ which generate $B_0$ as an $A_0$-module. Set $h_0 = \sum _{j = 1, \ldots , d} x_ jb_{0, j}$ in $B_0[x_1, \ldots , x_ d]$. Denote $h$, resp. $h_ i$ the image of $h_0$ in $B[x_1, \ldots , x_ d]$, resp. $B_ i[x_1, \ldots , x_ d]$. The image $U$ of $D(h)$ in $\mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ d])$ is open as $A \to B$ is universally open. Of course $U$ is quasi-compact as the image of an affine scheme. For $i$ large enough there is a quasi-compact open $U_ i \subset \mathop{\mathrm{Spec}}(A_ i[x_1, \ldots , x_ d])$ whose inverse image in $\mathop{\mathrm{Spec}}(A[x_1, \ldots , x_ d])$ is $U$, see Limits, Lemma 32.4.11. After increasing $i$ we may assume that $D(h_ i)$ maps into $U_ i$; this follows from the same lemma by considering the pullback of $U_ i$ in $D(h_ i)$. Finally, for $i$ even larger the morphism of schemes $D(h_ i) \to U_ i$ will be surjective by an application of the already used Limits, Lemma 32.8.15. We conclude $A_ i \to B_ i$ is universally open by Lemma 37.74.3. $\square$

Proof. Agreement of the functions is immediate from the fact that the (geometric) fibres of a locally quasi-finite morphism are discrete, see Morphisms, Lemma 29.20.8. Boundedness follows from Morphisms, Lemmas 29.57.2 and 29.57.9. $\square$

Proof. The question is local on $Y$ hence we may assume $Y$ affine. Let $K$ be an algebraic closure of the residue field $\kappa (y)$. Our assumption is that $(X_ y)_ K$ has $\geq d$ connected components. Then for a suitable quasi-compact open $X' \subset X$ the scheme $(X'_ y)_ K$ has $\geq d$ connected components; details omitted. After replacing $X$ by $X'$ we may assume $X$ is quasi-compact. Then $f$ is quasi-finite. Let $x_1, \ldots , x_ n$ be the points of $X$ lying over $y$. Apply Lemma 37.41.5 to get an étale neighbourhood $(U, u) \to (Y, y)$ and a decomposition

as in locus citatus. Observe that $n_{X/Y}(y) = \sum _ i m_ i$ in this situation; some details omitted. Since $f$ is universally open, we see that $V_{i, j} \to U$ is open for all $i, j$. Hence after shrinking $U$ we may assume $V_{i, j} \to U$ is surjective for all $i, j$. This proves that $n_{U \times _ Y X/U} \geq \sum _ i m_ i = n_{X/Y}(y) \geq d$. Since the construction of $n_{X/Y}$ is compatible with base change the proof is complete. $\square$

Proof. The openness of $Y_ d$ is immediate from Lemma 37.74.6. To prove finiteness over $Y_ d$ we redo the argument of the proof of that lemma. Namely, let $y \in Y_ d$. Then there are at most $d$ points of $X$ lying over $y$. Say $x_1, \ldots , x_ n$ are the points of $X$ lying over $y$. Apply Lemma 37.41.5 to get an étale neighbourhood $(U, u) \to (Y, y)$ and a decomposition

as in locus citatus. Observe that $d = n_{X/Y}(y) = \sum _ i m_ i$ in this situation; some details omitted. Since $f$ is universally open, we see that $V_{i, j} \to U$ is open for all $i, j$. Hence after shrinking $U$ we may assume $V_{i, j} \to U$ is surjective for all $i, j$ and we may assume $U$ maps into $W$. This proves that $n_{U \times _ Y X/U} \geq \sum _ i m_ i = d$. Since the construction of $n_{X/Y}$ is compatible with base change we know that $n_{U \times _ Y X/U} = d$. This means that $W$ has to be empty and we conclude that $U \times _ Y X \to U$ is finite. By Descent, Lemma 35.23.23 this implies that $X \to Y$ is finite over the image of the open morphism $U \to Y$. In other words, we see that $f$ is finite over an open neighbourhood of $y$ as desired. $\square$