Éléments de Géométrie Algébrique — English

§14. Universally open morphisms

§§14 and 15 are devoted to the study of the notion of universally open morphism (2.4.2). One has already seen (2.4.6) that a flat morphism locally of finite presentation is universally open, the converse being inexact. In §14 we first examine the properties of the dimensions of the fibres of a universally open morphism $f:X\to Y$; when $X$ and $Y$ are locally Noetherian, $f$ behaves in this respect (14.2.1) like a flat morphism (cf. (6.1.2)), and is in particular equidimensional when it is locally of finite type and dominant and $X$ is irreducible (14.2.2). Conversely, an equidimensional morphism $f:X\to Y$ locally of finite type is universally open when one supposes in addition that $Y$ is geometrically unibranch, and in particular when $Y$ is normal (Chevalley's criterion, (14.4.4)). We show also that the universally open morphisms $f:X\to Y$ locally of finite type (when $X$ and $Y$ are locally Noetherian and $Y$ is irreducible) admit "sufficiently many" quasi-sections, i.e. in a neighbourhood of a closed point $x$ of a fibre $f^{-1}(y)$, there exists a closed subprescheme $X^{\prime }$ of $X$ containing $x$ such that the restriction $X^{\prime }\to Y$ of $f$ is a quasi-finite (hence with discrete fibres) and dominant morphism (14.5.3).

In §15 we study various properties of the fibres of universally open morphisms $f:X\to Y$ locally of finite type, notably when $X$ and $Y$ are locally Noetherian. One thus obtains in particular a criterion for a point $x\in X$ to belong to only one irreducible component of $X$, in terms of properties of the fibre $f^{-1}(f(x))$ of $x$: it suffices that $Y$ be geometrically unibranch (for example normal) at the point $f(x)$ and that $f^{-1}(f(x))$ be geometrically pointwise integral at the point $x$ (15.3.3); if in addition $Y$ is locally integral at the point $f(x)$, $f$ is flat at the point $x$ and $X$ is locally integral at the point $x$. We also study the variation of the geometric number of connected components of a fibre $f^{-1}(y)$; for example, if $f$ is universally open and proper, and the fibres of $f$ are geometrically reduced, this number is locally constant (15.5.7). Finally, when $f$ admits a section $g$ (which will be the case when $X$ is a $Y$-group scheme), and when for every $y\in Y$ one denotes by $X_y^{\circ }$ the connected component of the fibre $X_y=f^{-1}(y)$ at the point $g(y)$ (the "neutral component" in the case of groups), one studies the union $X^{\circ }$ of the $X_y^{\circ }$ for $y\in Y$, and one shows (15.6.4) that if $f$ is universally open and the fibres $X_y$ geometrically reduced, then $X^{\circ }$ is an open set in $X$.

14.1. Open morphisms

(14.1.1)

Recall (1.10.2) that a continuous map $\psi :X\to Y$ is said to be open at a point $x\in X$ if the image under $\psi$ of every neighbourhood of $x$ in $X$ is a neighbourhood of $\psi (x)$ in $Y$.

One notes that this does not imply that there exists a fundamental system of neighbourhoods of $x$ whose images are open in $Y$.

Proposition (14.1.2).

Let $X$, $Y$ be two topological spaces, $\psi :X\to Y$ a continuous map, $x$ a point of $X$, $y=\psi (x)$.

(i) If $\psi$ is open at $x$, then for every part $Y^{\prime }$ of $Y$ containing $\psi (x)$ the restriction ${\psi }^{-1}(Y^{\prime })\to Y^{\prime }$ of $\psi$ to $Y^{\prime }$ is open at the point $x$.

(ii) Suppose that $Y$ is the union of a locally finite family of closed parts $(Y_i)$ and that for every $i$ such that $\psi (x)\in Y_i$, the restriction ${\psi }^{-1}(Y_i)\to Y_i$ of $\psi$ is open at the point $x$; then $\psi$ is open at the point $x$.

(iii) Let $\gamma :X^{\prime }\to X$ be a continuous map, $x^{\prime }$ a point of $X^{\prime }$; if the composite map $\psi \circ \gamma :X^{\prime }\to Y$ is open at the point $x^{\prime }$, then $\psi$ is open at the point $\gamma (x^{\prime })$.

If $U$ is a neighbourhood of $x$, one has $\psi (U\cap {\psi }^{-1}(Y^{\prime }))=\psi (U)\cap Y^{\prime }$; whence (i) at once. To prove (ii), note that there is a neighbourhood $W$ of $\psi (x)$ in $Y$ meeting only finitely many of the closed parts $Y_i$ that contain $\psi (x)$; hence $W$ is the union of the $W\cap Y_i$ for these indices. Now, if $U$ is a neighbourhood of $x$ such that $\psi (U)\cap Y_i$ is a neighbourhood of $\psi (x)$ in $Y_i$, there exists a neighbourhood $V\subset W$ of $\psi (x)$ in $Y$ such that for all $i$ with $\psi (x)\in Y_i$ one has $V\cap Y_i\subset \psi (U)\cap Y_i$, and since the union of the $V\cap Y_i$ for these indices is $V$, one has $V\subset \psi (U)$, hence $\psi (U)$ is a neighbourhood of $\psi (x)$ in $Y$. Assertion (iii) is trivial.

Remarks (14.1.3).

(i) The set $Z$ of points $x\in X$ where a morphism is open is not necessarily open. For example, let $K$ be a field, $A$ the polynomial ring K[S, T], $V$ the affine plane $\mathrm{Spec}(A)$, $X$ the closed subprescheme of $V$ "the union of the line X_1 defined by $T=0$ and the line X_2 defined by $S=0$", that is to say $\mathrm{Spec}(A/\mathfrak{a})$, where $\mathfrak{a}=AST$; take $Y=X_2=\mathrm{Spec}(A/AS)$ and for $f$ the projection corresponding to the canonical injection $K[T]\to A/\mathfrak{a}$; then one has $Z=X_2$, which is not open in $X$.

(ii) Let $X$, $Y$ be two Noetherian irreducible preschemes with generic points $\xi$, $\eta$ respectively, and $f:X\to Y$ a dominant morphism locally of finite type; then $f$ is open at the point $\xi$. Indeed, one may obviously restrict to the case where $X$ and $Y$ are reduced (hence integral) (1.5.4) and affine; by virtue of the generic flatness theorem (6.9.1), there exists a non-empty open set $V$ of $Y$ such that the restriction $f^{-1}(V)\to V$ of $f$ is a flat morphism, and one concludes from (2.4.6) that this restriction is an open morphism. However, as there exist dominant morphisms of finite type $f:X\to Y$ (where $X$ and $Y$ are irreducible) which are not open (see for example (II, 8.1)), the set of points where such a morphism is open is not necessarily closed in $X$.

(iii) We do not know whether, when $X$ and $Y$ are locally Noetherian and $f:X\to Y$ is a morphism locally of finite type, the set of points of $X$ where $f$ is open is or is not locally constructible.

Proposition (14.1.4).

Let $X$, $Y$ be two topological spaces, $\psi :X\to Y$ a continuous map. For every $y\in Y$, the set of $x\in {\psi }^{-1}(y)$ where $\psi$ is open is a closed part of ${\psi }^{-1}(y)$.

Indeed, suppose that $\psi$ is not open at a point $x\in {\psi }^{-1}(y)$; there exists then an open neighbourhood $V$ of $x$ in $X$ such that $\psi (V)$ is not a neighbourhood of $y$; it follows that for every $x^{\prime }\in V\cap {\psi }^{-1}(y)$, $\psi$ is not open at the point $x^{\prime }$.

Remark (14.1.5).

Even if $X$ and $Y$ are locally Noetherian and $f$ is a finite morphism, it can happen that $f$ be open at all points of a fibre $f^{-1}(y)$, without there existing a neighbourhood of $f^{-1}(y)$ at all points of which $f$ is open. Let for example $K$ be a field, $A$ the polynomial ring $K[T_1,T_2,T_3]$, $V=\mathrm{Spec}(A)$ affine 3-space over $K$, $X=\mathrm{Spec}(A/\mathfrak{a})$, where $\mathfrak{a}=\mathfrak{bc}$, with $\mathfrak{b}=(T_3)$ and $\mathfrak{c}=(T_1)+(T_2-T_3)$ in $A$, so that $X$ is the union of the plane $X_1=\mathrm{Spec}(A/\mathfrak{b})$ ("plane of equation $T_3=0$") and the line $X_2=\mathrm{Spec}(A/\mathfrak{c})$ ("line of equations $T_1=0$, $T_2=T_3$"), which are its irreducible components. Take $Y=X_1$ and let $f:X\to Y$ be the projection corresponding to the canonical injection $K[T_1,T_2]\to A/\mathfrak{a}$; if $y$ is the point common to X_1 and X_2, $f^{-1}(y)$ reduces to $y$ and $f$ is open at this point but is open at no point of X_2 in a neighbourhood of $y$ and distinct from $y$.

(14.1.6)

In what follows, the essential role will be played by the criterion (1.10.3) characterizing the morphisms locally of finite presentation $f:X\to Y$ that are open at a point $x$ by "lifting of generizations": for every generization $y^{\prime }$ of $y=f(x)$, there exists $x^{\prime }\in X$, a generization of $x$, such that $y^{\prime }=f(x^{\prime })$.

14.2. Open morphisms and the dimension formula

Theorem (14.2.1).

Let $X$, $Y$ be two locally Noetherian preschemes, $f:X\to Y$ a morphism, $x$ a point of $X$, $y=f(x)$. Suppose that $f$ is open at the generic points of the irreducible components of $f^{-1}(y)$ containing $x$. Then one has the relation

  (14.2.1.1)             dim(𝒪_x) = dim(𝒪_y) + dim_x(f⁻¹(y)).

Let $y^{\prime }$ be any generization of $y$ distinct from $y$, and consider the reduced closed subprescheme $Y^{\prime }$ of $Y$ with underlying space $\overline{y^{\prime }}$; then no irreducible component $X^{\prime }$ of $f^{-1}(Y^{\prime })$ containing $x$ can be contained in $f^{-1}(y)$: indeed, if $x^{\prime }$ is the generic point of $X^{\prime }$, one would have $f(x^{\prime })=y\ne y^{\prime }$, and as $x^{\prime }$ is its only generization in $f^{-1}(Y^{\prime })$, this would contradict the hypothesis that $f$ is open at the point $x^{\prime }$, by virtue of the fact that a) implies c) in (1.10.3). One can therefore apply (6.1.2) to the local homomorphism of Noetherian rings ${\mathcal{O}}_y\to {\mathcal{O}}_x$, whence the conclusion.

Corollary (14.2.2).

Let $Y$ be a locally Noetherian prescheme, $f:X\to Y$ a morphism locally of finite type, $x$ a point of $X$, $y=f(x)$. Suppose $f$ open at the generic points of the irreducible components of $f^{-1}(y)$ containing $x$. Then $f$ is equidimensional at the point $x$ in each of the following two cases:

(i) $X$ is irreducible and $f$ is dominant.

(ii) The rings ${\mathcal{O}}_x$ and ${\mathcal{O}}_y$ are equidimensional.

For (i), this results from (14.2.1) and (13.2.3). For (ii), this results from (14.2.1) and (13.3.6).

Proposition (14.2.3).

Let $Y$ be an irreducible locally Noetherian prescheme, $\eta$ its generic point, $f:X\to Y$ a morphism locally of finite type, $y$ a point of $f(X)$, $Z$ an irreducible component of $f^{-1}(y)$ such that $f$ is open at the generic point of $Z$. Then $Z$ is contained in an irreducible component $X^{\prime }$ of $X$ dominating $Y$ and such that

  dim_z(X') = dim(𝒪_y) + dim(Z) = dim(𝒪_y) + dim_z(f⁻¹(y)).

This results indeed from (14.2.1) applied to the generic point of $Z$, and from (13.2.7).

Corollary (14.2.4).

With the notation of (14.2.3), suppose that $f$ is open at all points of $f^{-1}(y)$. For every $z\in Y$, let $E(z)$ be the set of dimensions of the irreducible components of $f^{-1}(z)$ and $d(z)=\sup E(z)=\dim (f^{-1}(z))$. Then one has $E(y)\subset E(\eta )$, whence $d(y)\le d(\eta )$.

This results at once from (14.2.3).

Corollary (14.2.5).

Let $Y$ be a locally Noetherian prescheme, $f:X\to Y$ a proper morphism, $y\in Y$ a point such that $f$ is open at all points of $f^{-1}(y)$. Then the function $z\mapsto \dim (f^{-1}(z))$ is constant in a neighbourhood of $y$.

Let $Y_i$ ($1\le i\le n$) be the reduced closed subpreschemes of $Y$ whose underlying spaces are the irreducible components of $Y$ containing $y$; if $f_i:f^{-1}(Y_i)\to Y_i$ is the restriction of $f$, one knows that each of the $f_i$ is proper (II, 5.4.5) and open at all points of $f^{-1}(y)$ (14.1.2). As the union of the $Y_i$ is a neighbourhood of $y$ in $Y$,

one sees that one may restrict to proving the corollary when $Y$ is irreducible; let $\eta$ be its generic point. It follows from (14.2.4) that $d(y)\le d(\eta )$; on the other hand, since $f$ is proper, one deduces from (13.1.5) that $z\mapsto \dim (f^{-1}(z))$ is upper semi-continuous; in particular $d(y)\ge d(\eta )$, hence $d(y)=d(\eta )$. Moreover, there is a neighbourhood $V$ of $y$ such that $d(z)\le d(y)$ for every $z\in V$; but as $z$ is a specialization of $\eta$, one has on the other hand $d(z)\ge d(\eta )$, whence finally $d(z)=d(y)$ for $z\in V$.

Corollary (14.2.6).

Let $Y$ be a locally Noetherian prescheme, $f:X\to Y$ an open morphism of finite type, $y$ a point of $Y$. Let $y_i$ ($1\le i\le n$) be the generic points of the irreducible components of $Y$ containing $y$. Then, with the notation of (14.2.4):

(i) If for $1\le i\le n$, one has $E(y)=E(y_i)$ (resp. $d(y)=d(y_i)$), the function $E$ (resp. $d$) is constant in a neighbourhood of $y$.

(ii) There exists an open neighbourhood $U$ of $f^{-1}(y)$ such that the function $z\mapsto \dim (U\cap f^{-1}(z))$ is constant in a neighbourhood of $y$.

(i) The same reasoning as in (14.2.5) shows that one may restrict to the case where $Y$ is irreducible, of generic point $\eta$. If $y^{\prime }$ is any generization of $y$, one has (applying (14.2.4) to the restriction $f^{-1}(Y^{\prime })\to Y^{\prime }$ of $f$, where $Y^{\prime }=\overline{y^{\prime }}$, and using (14.1.2)) $E(y)\subset E(y^{\prime })\subset E(\eta )$ and $d(y)\le d(y^{\prime })\le d(\eta )$; this shows that the relation $E(y)=E(\eta )$ (resp. $d(y)=d(\eta )$) entails $E(y)=E(y^{\prime })$ (resp. $d(y)=d(y^{\prime })$) for every generization $y^{\prime }$ of $y$. Now, by virtue of (9.5.5), the functions $E$ and $d$ are locally constructible, hence it follows from $(0_{III},\mathrm{9.2.5})$ applied to the set of $z$ such that $E(z)=E(y)$ (resp. $d(z)=d(y)$) that this set is a neighbourhood of $y$.

(ii) For each of the $y_i$, $f^{-1}(y_i)$ is a Noetherian space since $f$ is of finite type; let $x_{ij}$ ($1\le j\le m_i$) be the generic points of its irreducible components, and set $X_{ij}=\overline{x_{ij}}$; we show that the complement $U$ in $X$ of the union of the $X_{ij}$ that do not meet $f^{-1}(y)$ (which is evidently an open neighbourhood of $f^{-1}(y)$) answers the question. Indeed, the restriction $f|U:U\to Y$ of $f$ is an open morphism of finite type (I, 6.3.5); for every pair $(i,j)$ such that $x_{ij}\in U$, $x_{ij}$ is a maximal point of $U\cap f^{-1}(y_i)$ and it follows from (13.1.1) applied to the restriction $X_{ij}\to \overline{y_i}$ of $f$ (taking into account (I, 5.2.2) and (I, 5.4)) that all the irreducible components of $X_{ij}\cap f^{-1}(y)$ have a dimension $\ge \dim (X_{ij}\cap f^{-1}(y_i))$. Taking into account (14.2.3) applied to the restriction $f^{-1}(Y_i)\to Y_i$ of $f$ which is an open morphism (14.1.2), $f^{-1}(y)$ is, for each $i$, the union of the irreducible components of $X_{ij}\cap f^{-1}(y)$ ($1\le j\le m_i$), hence $\dim (f^{-1}(y))\ge \dim (U\cap f^{-1}(y))$; but since $f|U$ is open, one also has $\dim (f^{-1}(y))\le \dim (f^{-1}(y)\cap U)$ by virtue of (14.2.4); one sees therefore that one may apply (i) to $f|U$, whence the conclusion.

Remark (14.2.7).

The example (13.2.12, (iii)) shows that under the hypotheses of (14.2.5) or (14.2.6, (ii)), one cannot, in the conclusion, replace the function $z\mapsto \dim (f^{-1}(z))$ by the function $z\mapsto E(z)$; indeed, in that example, $f$ is proper and flat (hence universally open (2.4.6)) and every neighbourhood of the unique point of $f^{-1}(y)$ contains the generic points of the irreducible components of $f^{-1}(\eta )$.

14.3. Universally open morphisms

(14.3.1)

Recall (2.4.2) that to say that a morphism $f:X\to Y$ is universally open signifies that for every morphism $g:Y^{\prime }\to Y$, $f_{(Y^{\prime })}:X_{(Y^{\prime })}\to Y^{\prime }$ is open; it moreover suffices that this hold when $Y^{\prime }=Y[T_1,\cdots ,T_e]$, for every $e$ (8.10.2) (and if $Y$ is locally Noetherian, it therefore suffices that this hold for every locally Noetherian $Y^{\prime }$).

Proposition (14.3.2).

Let $f:X\to Y$ be a morphism of preschemes. If $f$ is universally open, then, for every morphism $g:Y^{\prime }\to Y$, where $Y^{\prime }$ is irreducible, the image under $f^{\prime }=f_{(Y^{\prime })}:X_{(Y^{\prime })}\to Y^{\prime }$ of every irreducible component of $X_{(Y^{\prime })}$ is dense in $Y^{\prime }$. Conversely, if this condition is satisfied for every irreducible $Y^{\prime }$ and every morphism of finite type $g:Y^{\prime }\to Y$, and if moreover $f$ is locally of finite presentation, then $f$ is universally open.

Indeed, it follows from (1.10.4) that if $f$ is universally open, it satisfies the condition of the statement. Conversely, suppose that $f$ is locally of finite presentation, and let us show that for every integer $e$, if one sets $Y^{\prime \prime }=Y[T_1,\cdots ,T_e]$, $f_{(Y^{\prime \prime })}$ is open. Indeed, let $Y^{\prime }$ be a closed subprescheme of Y'' whose underlying space is an irreducible closed part of Y''; the composite morphism $Y^{\prime }\to Y^{\prime \prime }\to Y$ is of finite type, hence every irreducible component of $X_{(Y^{\prime })}$ dominates $Y^{\prime }$ by hypothesis; one therefore deduces from (1.10.4) that $f_{(Y^{\prime \prime })}$ is an open morphism.

This proposition shows that the definition (III, 4.3.9) coincides in the case considered with the general definition of universally open morphisms given in (2.4.2).

To the notion of morphism open at a point (14.1.1) likewise corresponds the following:

Definition (14.3.3).

Let $f:X\to Y$ be a morphism of preschemes, $x$ a point of $X$. One says that $f$ is universally open at the point $x$ if, for every morphism $g:Y^{\prime }\to Y$, setting $X^{\prime }=X{\times }_Y{Y}^{\prime }$, the morphism $f^{\prime }=f_{(Y^{\prime })}:X^{\prime }\to Y^{\prime }$ is open at every point $x^{\prime }$ of $X^{\prime }$ whose projection in $X$ is $x$.

Remarks (14.3.3.1).

(i) The reasoning of (8.10.1) shows (with the same notation) that if $x$ is a point of $X$ and $x_{\lambda }$ its projection in $X_{\lambda }$, then, if $f_{\lambda }$ is open at the point $x_{\lambda }$ for every $\lambda \in X$, $f$ is open at the point $x$; it suffices to restrict to the open sets $U$ of $X$ containing $x$ and to remark that the hypothesis implies that $f_{\lambda }(U_{\lambda })$ is a neighbourhood of $f_{\lambda }(x_{\lambda })$, hence $f(p_{\lambda }^{-1}(U_{\lambda }))$ is a neighbourhood of $f(x)$. One deduces that the statement (8.10.2) is still exact when one replaces "universally open" by "universally open at the point $x$", and "open morphism" by "open morphism at every point x'' of X'' whose projection in $X$ is $x$": it suffices in the proof to restrict to the open sets $V$ containing some x''.

(ii) The result of (14.1.4) remains valid for a morphism $f:X\to Y$, replacing "open" by "universally open". Indeed, suppose that $f$ is not universally open at a point $x\in f^{-1}(y)$; there is consequently a morphism $Y^{\prime }\to Y$ and a point $x^{\prime }\in X^{\prime }$ projecting to $x$, such that $f^{\prime }$ is not open at the point $x^{\prime }$.

Now, if $y^{\prime }=f^{\prime }(x^{\prime })$, the projection $f^{\prime -1}(y^{\prime })\to f^{-1}(y)$ is an open morphism (2.4.10), and there is, by virtue of (14.1.4), a neighbourhood $V$ of $x^{\prime }$ in $f^{\prime -1}(y^{\prime })$ where $f^{\prime }$ is not open; hence $f$ is not universally open at the points of the image of $V$ in $f^{-1}(y)$, which is a neighbourhood of $x$ in $f^{-1}(y)$.

Proposition (14.3.4).

(i) Let $f:X\to Y$, $g:Y\to Z$ be two morphisms, $x$ a point of $X$, $y=f(x)$. If $f$ is universally open at the point $x$ and $g$ universally open at the point $y$, then $g\circ f$ is universally open at the point $x$. Conversely, if $g\circ f$ is universally open at the point $x$, $g$ is universally open at the point $y$.

(ii) If $f:X\to Y$ is an $S$-morphism universally open at the point $x\in X$, then, for every base change $S^{\prime }\to S$, $f_{(S^{\prime })}:X_{(S^{\prime })}\to Y_{(S^{\prime })}$ is universally open at every point of $X_{(S^{\prime })}$ above $x$.

(iii) For $f:X\to Y$ to be universally open at the point $x\in X$, it is necessary and sufficient that $f_{red}$ be so.

(iv) Let $f:X\to Y$ be a morphism locally of finite presentation, $x$ a point of $X$, $y=f(x)$; set $Y_1=\mathrm{Spec}({\mathcal{O}}_y)$, $X_1=X{\times }_Y{Y}_1$, $f_1=f_{(Y_1)}$. For $f$ to be universally open at the point $x$, it is necessary and sufficient that $f_1$ be so (one recalls (I, 3.6.5) that X_1 is canonically identified with a subspace of $X$).

Indeed (ii) is an evident consequence of the definition (14.3.3); it also results from the definition that to prove assertion (i), it suffices to do so when one suppresses the word "universally" everywhere, and this then results from (14.1.2). Assertion (iii) results from (ii) and from the fact that the canonical morphism $X_{red}\to X$ is surjective. Finally, condition (iv) is trivially necessary. On the other hand, if it is satisfied, and if $g:Y^{\prime }\to Y$ is an arbitrary morphism, $X^{\prime }=X_{(Y^{\prime })}$, $f^{\prime }=f_{(Y^{\prime })}$, $x^{\prime }$ a point of $X^{\prime }$ above $x$, then $f^{\prime }$ is locally of finite presentation, and to see that it is open at the point $x^{\prime }$, it suffices to apply the criterion (1.10.3, c)). Set $y^{\prime }=f^{\prime }(x^{\prime })$, $Y_1^{\prime }=\mathrm{Spec}({\mathcal{O}}_{y^{\prime }})$, $X_1^{\prime }=X^{\prime }{\times }_{Y^{\prime }}{Y}_1^{\prime }$, $f_1^{\prime }=f_{(Y_1^{\prime })}^{\prime }$. Since $g(y^{\prime })=y$, the composite morphism $Y_1^{\prime }\to Y^{\prime }\to Y$ factors as $Y_1^{\prime }\to Y_1\to Y$ (I, 2.4.4), hence $X_1^{\prime }=X_1{\times }_{Y_1}{Y}_1^{\prime }$ and $f_1^{\prime }=(f_1{)}_{(Y_1^{\prime })}$; the conclusion then results at once from the hypothesis that $f_1$ is open at the point $x^{\prime }$ and from (1.10.3).

Proposition (14.3.5).

Let $X$, $Y$ be two preschemes, $f:X\to Y$ a morphism, $x$ a point of $X$. Let $(Y_i{)}_{1\le i\le n}$ be a locally finite family of closed subpreschemes of $Y$ such that the space $Y$ is the union of the $Y_i$, and suppose that for every $i$ such that $f(x)\in Y_i$, the restriction $f^{-1}(Y_i)\to Y_i$ of $f$ is a morphism universally open at the point $x$; then $f$ is universally open at the point $x$.

Taking the definition into account, this results from the analogous proposition (14.1.2, (ii)) for morphisms open at a point.

Proposition (14.3.6).

Let $Y$ be a locally Noetherian prescheme, $f:X\to Y$ a morphism locally of finite type. For $f$ to be universally open at a point $x\in X$, it is necessary and sufficient that the following condition be satisfied: for every morphism $g:Y^{\prime }\to Y$, where $Y^{\prime }=\mathrm{Spec}(A)$ is the spectrum of a discrete valuation ring, such that the image $g(y^{\prime })$ of the closed point $y^{\prime }$ of $Y^{\prime }$ is equal to $y=f(x)$, and for every point $x^{\prime }\in X^{\prime }=X{\times }_Y{Y}^{\prime }$ whose projections on $X$ and $Y^{\prime }$ are $x$ and $y^{\prime }$ respectively, there exists a generization $z^{\prime }$ of $x^{\prime }$ in $X^{\prime }$ whose projection in $Y^{\prime }$ is the generic point of $Y^{\prime }$ (in other words, there exists an irreducible component of $X^{\prime }$ containing $x^{\prime }$ and dominating $Y^{\prime }$).

Moreover, one may, in the preceding condition, restrict to the case where $A$ is complete, has an algebraically closed residue field, and where $x^{\prime }$ is rational over $k(y^{\prime })$.

If $Y^{\prime }$ is as in the statement, the necessity of the condition results from the fact that $f^{\prime }$ must be open at the point $x^{\prime }$, and from the criterion (1.10.3). To see that the condition is sufficient, consider a morphism of finite type $g:Y^{\prime \prime }\to Y$, and let $X^{\prime \prime }=X{\times }_Y{Y}^{\prime \prime }$, $f^{\prime \prime }=f_{(Y^{\prime \prime })}:X^{\prime \prime }\to Y^{\prime \prime }$, and x'' a point of X'' above $x$. Set $y^{\prime \prime }=f^{\prime \prime }(x^{\prime \prime })$, and let $t$ be a generization of y'' in Y'', distinct from y''. Since Y'' is locally Noetherian, it follows from (II, 7.1.9) and $(0_{III},\mathrm{10.3.1})$ that there exists a scheme $Y^{\prime }=\mathrm{Spec}(A)$, where $A$ is a complete discrete valuation ring whose residue field is an algebraic closure of $k(x^{\prime \prime })$, and a morphism $g:Y^{\prime }\to Y^{\prime \prime }$ such that, if $s^{\prime }$ and $y^{\prime }$ are the generic point and closed point of $Y^{\prime }$, one has $g(s^{\prime })=t$ and $g(y^{\prime })=y^{\prime \prime }$. There is then a point $x^{\prime }$ of $X^{\prime }=X{\times }_Y{Y}^{\prime }=X^{\prime \prime }{\times }_{Y^{\prime \prime }}{Y}^{\prime }$ whose projections in X'' and $Y^{\prime }$ are x'' and $y^{\prime }$ respectively, and which is rational over $k(y^{\prime })$ (I, 3.4.9). The hypothesis implies that there is a generization $z^{\prime }$ of $x^{\prime }$ in $X^{\prime }$ whose projection in $Y^{\prime }$ is $s^{\prime }$; if z'' is the projection of $z^{\prime }$ in X'', z'' is a generization of x'' and its projection in Y'' is $t$; one therefore concludes from (1.10.3) that f'' is open at the point x'', hence that $f$ is universally open at the point $x$ (14.3.3.1, (i)).

Corollary (14.3.7).

The notation being that of (14.3.6):

(i) Given a point $y\in Y$, for $f$ to be universally open at all points of $f^{-1}(y)$, it is necessary and sufficient that for every morphism $g:Y^{\prime }\to Y$, where $Y^{\prime }$ is the spectrum of a discrete valuation ring, and where the image $g(y^{\prime })$ of the closed point $y^{\prime }$ of $Y^{\prime }$ is $y$, every irreducible component of $X^{\prime }=X{\times }_Y{Y}^{\prime }$ dominates $Y^{\prime }$.

(ii) For $f$ to be universally open, it is necessary and sufficient that for every morphism $g:Y^{\prime }\to Y$, where $Y^{\prime }$ is the spectrum of a discrete valuation ring, every irreducible component of $X^{\prime }=X{\times }_Y{Y}^{\prime }$ dominates $Y^{\prime }$.

It is clear that it suffices to prove (i); the necessity of (i) results from (14.3.2) and its sufficiency from (14.3.6).

Proposition (14.3.8).

Let $Y$ be a locally Noetherian prescheme, irreducible, regular and of dimension 1 (for example the spectrum of a Dedekind ring), $f:X\to Y$ a morphism locally of finite type, $y$ a point of $Y$. The following conditions are equivalent:

a) $f_{red}$ is flat at every point of $f^{-1}(y)$.

b) $f$ is universally open in a neighbourhood of $f^{-1}(y)$.

c) $f$ is open in a neighbourhood of $f^{-1}(y)$.

d) Every irreducible component of $X$ meeting $f^{-1}(y)$ dominates $Y$.

Since $f_{red}$ is locally of finite type (1.3.4), a) entails that $f_{red}$ is flat in a neighbourhood of $f^{-1}(y)$ (11.1.1), and it suffices to apply (2.4.6) in such a neighbourhood to see that a) entails b). The implication b) ⟹ c) is trivial, and the implication c) ⟹ d) results from (1.10.4) applied to a neighbourhood of $f^{-1}(y)$. It remains to see that d) entails a). One may obviously, by virtue of (1.3.4), restrict to the case where $X$ is reduced. The question being moreover local on $X$ and on $Y$, one may suppose $Y=\mathrm{Spec}(A)$ and $X=\mathrm{Spec}(B)$ affine; if $X_i$ ($1\le i\le n$) are the closed (integral) subpreschemes of $X$ whose

underlying spaces are the irreducible components of $X$, then, for every $x\in X$, ${\mathcal{O}}_{X,x}$, being reduced, is a sub-ring of the direct product of the ${\mathcal{O}}_{X_i,x}$; if $y=f(x)$, it will suffice to show that each of the ${\mathcal{O}}_{X_i,x}$ is a torsion-free ${\mathcal{O}}_y$-module, for it will then be the same for ${\mathcal{O}}_{X,x}$; as by hypothesis ${\mathcal{O}}_y$ is a regular local ring of dimension 1, that is to say (II, 7.1.6) a discrete valuation ring, it will then result from $(0_I,\mathrm{6.3.4})$ that ${\mathcal{O}}_{X,x}$ is a flat ${\mathcal{O}}_y$-module. But if $X_i=\mathrm{Spec}(B_i)$, where $B_i$ is an integral ring, hypothesis d) entails that the homomorphism $A\to B_i$ is injective (I, 1.2.7); hence $B_i$ is a torsion-free $A$-module, and a fortiori ${\mathcal{O}}_{X_i,x}$ is a torsion-free ${\mathcal{O}}_y$-module.

Remarks (14.3.9).

(i) In the statement of (14.3.8), one cannot dispense with the hypothesis that $Y$ is regular. With the notation of (11.7.5), take indeed $Y=\mathrm{Spec}(A)$, $X=\mathrm{Spec}(\hat{A})$, so that $f:X\to Y$ is a finite surjective morphism; as $A$ is an integral local ring of dimension 1, as is Â, it follows at once from (1.10.4) that the morphism $f$ is open. However $f$ is not universally open (nor a fortiori flat), as is shown by (11.7.5). One would have an analogous example by taking for $Y$ the local scheme at the double point of an algebraic curve having an "ordinary double point" and for $X$ the normalization of $Y$.

(ii) The example of (14.1.3, (i)) shows that the set of points of $X$ where a morphism is universally open is not necessarily open, the morphism $f$ in that example being universally open at all points where it is open. The example $f:X\to Y$ seen above in (i) shows similarly that the set of points where a morphism is universally open is not necessarily closed, for it is immediate that at all points of $X$ except one the morphism $f$ is universally open (it is even a local isomorphism). It would be interesting to know whether the set of points where a morphism is universally open is locally constructible.

The two following propositions have been pointed out to us by M. Artin:

Lemma (14.3.10).

Let $A$ be a valuation ring (not necessarily discrete), $k$ its residue field, $K$ its field of fractions, and set $S=\mathrm{Spec}(A)$. Let $X$ be an irreducible prescheme, $f:X\to S$ a dominant morphism of finite type; let $X_0=X{\otimes }_{A}k$ (resp. $X_1=X{\otimes }_{A}K$) be the fibre of $f$ at the closed point (resp. at the generic point) of $S$. Then, if $X_0\ne \varnothing$, one has $\dim (X_0)=\dim (X_1)$.

One may restrict to the case where $X$ is affine, replacing $X$ if need be by an affine open set containing a generic point of an irreducible component of X_0 of maximal dimension, and using (4.1.1.3). Let $n=\dim (X_0)\ge 0$; it follows from (13.3.1.1) that there exists a neighbourhood $U$ of X_0 in $X$ and an $S$-morphism quasi-finite $g:U\to S[T_1,\cdots ,T_n]=Z$ such that the restriction morphism $g_0:U_0=U\cap X_0\to Z_0=\mathrm{Spec}(k[T_1,\cdots ,T_n])$ is finite and surjective. By the base change $\mathrm{Spec}(K)\to S$ and restriction to the open set $U_1=U\cap X_1$ of X_1, one deduces from $g$ a quasi-finite morphism $g_1:U_1\to Z_1=\mathrm{Spec}(K[T_1,\cdots ,T_n])$. Since U_1 is dense in X_1, one has $\dim (U_1)=\dim (X_1)$ (4.1.1.3); the proposition will be established, by virtue of (4.1.2), if we prove that the morphism $g_1$ is dominant. Suppose the contrary; there would then exist a non-zero polynomial $F_1\in K[T_1,\cdots ,T_n]$ such that $g_1(U_1)\subset V(F_1)$. If $\omega$ is a valuation on $K$ associated with $A$, and if $(c_{\alpha })$ is the family of coefficients of F_1, one may, after multiplication of F_1 by a non-zero element of $K$, suppose that one has ${\inf }_{\alpha }(\omega (c_{\alpha }))=0$; in other words, F_1 comes from a polynomial $F\in A[T_1,\cdots ,T_n]$

(with which it identifies) such that the image F_0 of $F$ in $k[T_1,\cdots ,T_n]$ is non-zero. Consider then in $Z$ the closed set $V(F)$; one has $V(F)\cap Z_1=V(F_1)$, and as U_1 is dense in $U$ (since it contains the generic point of $X$), $g(U)\subset V(F)$; in

particular, one would have $g_0(U_0)\subset V(F)\cap Z_0=V(F_0)$; but since $F_0\ne 0$, $V(F_0)$ is a closed part of Z_0 distinct from Z_0, and one reaches a contradiction. Q.E.D.

Proposition (14.3.11).

Let $f:X\to S$ be a morphism of finite type, $(g_{\lambda }{)}_{\lambda \in L}$ a family of universally open morphisms $g_{\lambda }:Y_{\lambda }\to S$, and for every $\lambda \in L$, let $u_{\lambda }:Y_{\lambda }\to X$ be an $S$-morphism. For every $s\in S$, set $X_s=X{\times }_S\mathrm{Spec}(k(s))$, $(Y_{\lambda }{)}_s=Y_{\lambda }{\times }_S\mathrm{Spec}(k(s))$, and let $(u_{\lambda }{)}_s:(Y_{\lambda }{)}_s\to X_s$ be the morphism $u_{\lambda }{\times }_{S}1$ deduced from $u_{\lambda }$ by base change. Let $Z(s)$ be the closure in $X_s$ of the union of the sets $(u_{\lambda }{)}_s((Y_{\lambda }{)}_s)$, and set $d(s)=\dim (Z(s))$. Then, for every generization $s^{\prime }$ of a point $s\in S$, one has $d(s)\le d(s^{\prime })$.

One knows (II, 7.1.4) that there exists a valuation ring $A^{\prime }$ and a morphism $h:S^{\prime }=\mathrm{Spec}(A^{\prime })\to S$ such that if $a$ (resp. $b$) is the closed point (resp. the generic point) of $S^{\prime }$, one has $h(a)=s$, $h(b)=s^{\prime }$. Moreover, the projection morphism $p:X_s{\otimes }_{k(s)}k(a)\to X_s$ is surjective and open (2.4.10), hence makes $X_s$ a quotient space of $X_s{\otimes }_{k(s)}k(a)$ by an open equivalence relation; for every part $M$ of $X_s$, $p^{-1}(\bar{M})$ is therefore equal to the closure $\overline{p^{-1}(M)}$ (Bourbaki, Top. gén., chap. I, 4th ed., §5, n° 3, prop. 7); one reasons similarly for $X_{s^{\prime }}$, and taking into account (I, 3.4.8), (4.2.7) and the fact that the $g_{\lambda }$ are universally open, one sees that one may reduce to proving the proposition in the situation obtained after base change $S^{\prime }\to S$. Suppose therefore $S^{\prime }=S$, $s$ being the closed point and $s^{\prime }$ the generic point of $S$. The hypothesis that $g_{\lambda }$ is open entails that every irreducible component of $Y_{\lambda }$ dominates $S$ (1.10.4), hence that its generic point is a maximal point of $(Y_{\lambda }{)}_{s^{\prime }}$; if $Z$ denotes the closure in $X$ of $Z(s^{\prime })$, one therefore has $u_{\lambda }(Y_{\lambda })\subset Z$, and consequently $(u_{\lambda }{)}_s((Y_{\lambda }{)}_s)\subset Z_s=Z\cap X_s$; in other words, one has $Z(s)\subset Z_s$, whence $\dim (Z(s))\le \dim (Z_s)$. But applying (14.3.10) to a reduced subprescheme of $X$ whose underlying space is an irreducible component of $Z$, one obtains $\dim (Z_s)\le \dim (Z_{s^{\prime }})$ (the inequality coming from the fact that there may be irreducible components of $Z$ not meeting $X_s$). On the other hand, since $Z(s^{\prime })$ is by definition closed in $X_{s^{\prime }}$, one has $Z_{s^{\prime }}=Z(s^{\prime })$, which completes the proof.

Remark (14.3.12).

The case envisaged by M. Artin was that where $Y_{\lambda }=S$ for every $\lambda$, in other words the case where $(u_{\lambda })$ is a family of $S$-sections of $X$. Another useful case is that where the family $(u_{\lambda })$ is reduced to a single element; one can moreover always reduce to this case by considering the prescheme $Y$ sum of the $Y_{\lambda }$ and the morphisms $g:Y\to S$ and $u:Y\to X$ whose restrictions to each $Y_{\lambda }$ are respectively $g_{\lambda }$ and $u_{\lambda }$.

Proposition (14.3.13).

Let $f:X\to Y$ be a morphism locally of finite type, $y$ a point of $Y$, $x$ a maximal point of the fibre $X_y=X{\times }_Y\mathrm{Spec}(k(y))$.

Consider the following conditions:

a) $f$ is universally open at the point $x$ (or equivalently, at every point of the irreducible component of $X_y$ of generic point $x$ (14.3.3.1, (ii))).

b) For every irreducible component Y_0 of $Y$ containing $y$, there exists an irreducible component $Z$ of $X$ containing $x$, dominating Y_0 and such that dim_x(X_y) = dim_x(Z ∩ X_y) ≤ dim(Z ∩ X_η), where $\eta$ is the generic point of Y_0 (which entails that $Z$ is equidimensional over Y_0 at the point $x$ (13.2.2)).

b') For every open neighbourhood $U$ of $x$ in $X$ and every generization $y^{\prime }$ of $y$, one has $\dim (U\cap X_{y^{\prime }})\ge {\dim }_x(U\cap X_y)$.

Then one has the implications a) ⟹ b) ⟺ b').

To show that b) implies b'), it suffices to remark that $y^{\prime }$ belongs to an irreducible component Y_0 of $Y$ containing $y$, of generic point $\eta$; taking $Z$ as in b) and noting that the generic point of $Z$ (which is also that of $Z\cap X_{\eta }$ $(0_I,\mathrm{2.1.8})$) is contained in $U$, one has $\dim (U\cap X_{y^{\prime }})\ge \dim (U\cap Z\cap X_{y^{\prime }})$, and, by virtue of (13.1.6), $\dim (U\cap Z\cap X_{y^{\prime }})\ge \dim (Z\cap X_{\eta })$; whence the assertion, since ${\dim }_x(U\cap X_y)={\dim }_x(X_y)$ (4.1.1.3).

To prove that b') implies b), one may first replace $X$ by $f^{-1}(Y_0)$, hence suppose $Y_0=Y$; one may restrict to the case where $X$ and $Y$ are affine, since ${\dim }_x(U\cap X_y)={\dim }_x(X_y)$ (4.1.1.3). The irreducible components $W_i$ of the Noetherian prescheme $X_y$ are then finite in number, and the complement of the union of the ${\bar{W}}_i$ (closures in $X$) that do not contain $x$ is an open neighbourhood $V$ of $x$. Replacing $X$ by $V$, one may therefore suppose that $x\in {\bar{W}}_i$ for every $i$ (hypothesis b') entails $\dim (V\cap X_y)\ge 0$, hence $x$ belongs to one of the ${\bar{W}}_i$ for at least one $i$). Moreover, the ${\bar{W}}_i$ are exactly the irreducible components of $X$ that dominate $Y$ $(0_I,\mathrm{2.1.8})$; this being so, if one had, for each of these components $Z$, $\dim (Z\cap X_{\eta })<{\dim }_x(X_y)$, one would conclude $\dim (X_{\eta })<{\dim }_x(X_y)$, contrary to hypothesis b'). The relation ${\dim }_x(Z\cap X_y)=\dim (Z\cap X_{\eta })$ then results from (13.1.6).

It remains to prove that a) entails b'). Taking (II, 7.1.4) and the invariance of the hypotheses and the conclusion under base change into account (by virtue of (4.2.7)), one may restrict to the case where $Y$ is a spectrum of a valuation ring, with closed point $y$ and generic point $y^{\prime }$, and where $U=X$. The hypothesis that $f$ is open at the point $x$ entails that there exists an irreducible component $Z$ of $X$ containing $x$ and dominating $Y$ (1.10.3). Applying (14.3.10) to a neighbourhood of $x$ in $Z$, one concludes that $\dim (Z\cap X_{y^{\prime }})=\dim (Z\cap X_y)$; but since $x$ is maximal in $X_y$, $Z\cap X_y$ contains the irreducible component of $X_y$ of generic point $x$, hence ${\dim }_x(Z\cap X_y)={\dim }_x(X_y)$; on the other hand, one has $\dim (X_{y^{\prime }})\ge \dim (Z\cap X_{y^{\prime }})$, which completes the proof of b').

Remark (14.3.14).

We do not know whether in (14.3.13) the conclusion remains valid when one replaces the hypothesis a) by the weaker hypothesis that $f$ is open at the point $x$. One may show easily that it would suffice to treat the case where $Y$ is the spectrum of an integral local ring whose generic point is isolated, and where $X$ is a closed subprescheme of the vector bundle Y[T].

14.4. Chevalley's criterion for universally open morphisms

Theorem (14.4.1).

Let $f:X\to Y$ be a morphism locally of finite type, $y$ a point of $Y$, $x$ a maximal point of the fibre $X_y=X{\times }_Y\mathrm{Spec}(k(y))$. Suppose $y$ geometrically unibranch. Then the following conditions are equivalent:

a) $f$ is universally open at the point $x$ (or equivalently, at every point of the irreducible component of $X_y$ of generic point $x$ (14.3.3.1, (ii))).

b) If Y_0 is the unique irreducible component of $Y$ containing $y$ and $\eta$ its generic point, there exists an irreducible component $Z$ of $X$ containing $x$, dominating Y_0 and such that ${\dim }_x(Z\cap X_y)=\dim (Z\cap X_{\eta })$ (which signifies that $Z$ is equidimensional over Y_0 at the point $x$ (13.2.2)).

b') For every open neighbourhood $U$ of $x$ in $X$ and every generization $y^{\prime }$ of $y$, one has $\dim (U\cap X_{y^{\prime }})\ge {\dim }_x(U\cap X_y)$.

If moreover $Y$ is locally Noetherian, these conditions are also equivalent to the following:

c) $f$ is open at the point $x$.

Note first that since $({\mathcal{O}}_y{)}_{red}$ is integral, $y$ belongs to only one irreducible component Y_0 of $Y$. The fact that b) and b') are equivalent and that a) implies b') results from (14.3.13); on the other hand, if $Y$ is locally Noetherian, one has seen in (14.2.3) that c) implies b). It therefore remains to show that when $y$ is geometrically unibranch, b) entails a).

Lemma (14.4.1.1).

Let $Y$ be a prescheme, $Y^{\prime }=Y[T_1,\cdots ,T_n]$, $y^{\prime }$ a point of $Y^{\prime }$, $y$ its image in $Y$. If $Y$ is geometrically unibranch at the point $y$, then $Y^{\prime }$ is geometrically unibranch at the point $y^{\prime }$.

Indeed, since for every $y\in Y$, $\mathrm{Spec}(k(y)[T_1,\cdots ,T_n])$ is geometrically regular over $k(y)$ (0, 17.3.7), the structure morphism $Y^{\prime }\to Y$ is smooth (6.8.1), and it suffices to apply (11.3.14).

Lemma (14.4.1.2).

Let $A$ be an integral unibranch local ring, $B$ an integral ring containing $A$ and integral over $A$, $\mathfrak{n}$ a prime ideal of $B$ above the maximal ideal $\mathfrak{m}$ of $A$. Then the morphism $\mathrm{Spec}(B_{\mathfrak{n}})\to \mathrm{Spec}(A)$ is surjective; in other words, for every prime ideal $\mathfrak{p}$ of $A$, there exists a prime ideal $\mathfrak{q}$ of $B$ such that $\mathfrak{q}\subset \mathfrak{n}$ and $\mathfrak{q}\cap A=\mathfrak{p}$.

Let $K$ (resp. $L$) be the field of fractions of $A$ (resp. $B$), $A^{\prime }$ the integral closure of $A$, $B^{\prime }$ the sub-ring of $L$ generated by $A^{\prime }$ and $B$, so that one has a commutative diagram of canonical injections

                B  ─→  B'
                ↑       ↑
                A  ─→  A'

As $B^{\prime }$ is integral over $B$, there exists a prime ideal ${\mathfrak{n}}^{\prime }$ of $B^{\prime }$ such that ${\mathfrak{n}}^{\prime }\cap B=\mathfrak{n}$ (Bourbaki, Alg. comm., chap. V, §2, n° 1, th. 1), and (for the same reason) $\mathrm{Spec}(A^{\prime })\to \mathrm{Spec}(A)$ is surjective. On the other hand, since $A$ is unibranch, $A^{\prime }$ is a local ring; hence ${\mathfrak{n}}^{\prime }\cap A^{\prime }$, which is above the maximal ideal $\mathfrak{m}$ of $A$, is necessarily equal to the unique maximal ideal ${\mathfrak{m}}^{\prime }$ of $A^{\prime }$. By virtue of the second Cohen-Seidenberg theorem (loc. cit., §2, n° 4, th. 3), the morphism $\mathrm{Spec}(B_{{\mathfrak{n}}^{\prime }}^{\prime })\to \mathrm{Spec}(A^{\prime })$ is surjective, hence so is the composite Spec(B'_{𝔫'}) → Spec(A') → Spec(A); but this morphism is also the composite Spec(B'_{𝔫'}) → Spec(B_𝔫) → Spec(A), hence the morphism $\mathrm{Spec}(B_{\mathfrak{n}})\to \mathrm{Spec}(A)$ is surjective.

These lemmas being established, let us return to the proof of the implication b) ⟹ a) in (14.4.1). By virtue of (14.3.3.1, (i)), it suffices to prove that, for every integer $n\ge 0$ and every point $x^{\prime }$ of $X^{\prime }=X[T_1,\cdots ,T_n]$ above $x$, the morphism

$f^{\prime }:X^{\prime }\to Y^{\prime }=Y[T_1,\cdots ,T_n]$, deduced from $f$ by base change, is open at the point $x^{\prime }$. Taking the lemma (14.4.1.1), (2.3.4) and (4.2.7) into account, one is therefore reduced to proving that $f$ is open at the point $x$: moreover, it evidently suffices (14.1.2, (iii)) to show that the restriction of $f$ to a closed subprescheme of $X$ having $Z$ for underlying space is open at the point $x$, so that one may restrict to the case where $X=Z$ is irreducible. Replacing $X$ by an open neighbourhood $V$ of $x$ such that $V\cap X_y$ is irreducible, one may suppose, by virtue of (13.3.1), that the morphism $f$ factors as $X\to Y^{\prime \prime }=Y[T_1,\cdots ,T_m]\to Y$, where $g$ is quasi-finite, dominant and locally of finite type. As the structure morphism $Y^{\prime \prime }\to Y$ is open (2.4.6), one is reduced to proving that $g:X\to Y^{\prime \prime }$ is open at the point $x$. Moreover, by virtue of (14.4.1.1), $g(x)$ is a geometrically unibranch point of Y''. One is therefore reduced to proving the following lemma:

Lemma (14.4.1.3).

Let $X$, $Y$ be two irreducible preschemes, $f:X\to Y$ a morphism locally quasi-finite and dominant. If $x\in X$ is such that $y=f(x)$ is unibranch over $Y$, then $f$ is open at the point $x$.

It suffices to prove that $f(\mathrm{Spec}({\mathcal{O}}_{X,x}))=\mathrm{Spec}({\mathcal{O}}_{Y,y})$ (1.10.3). One may therefore restrict, by the base change $\mathrm{Spec}({\mathcal{O}}_{Y,y})\to Y$, to the case where $Y=\mathrm{Spec}(A)$, where $A$ is a local ring and $y$ is the closed point of $Y$ (taking into account (I, 3.6.5) and $(0_I,\mathrm{2.1.8})$, which prove that $X{\times }_Y\mathrm{Spec}({\mathcal{O}}_{Y,y})$ is irreducible); replacing $f$ by $f_{red}$, one may suppose $X$ and $Y$ reduced, hence integral. Replacing if necessary $X$ and $Y$ by affine neighbourhoods of $x$ and $y$ respectively, one may suppose (8.12.9) that the morphism $f$ factors as $X\to X_1\to Y$, where $j$ is an open immersion and $g$ a finite morphism (evidently dominant); as $X$ and X_1 are affine, $j$ is affine, hence separated and quasi-compact, and consequently factors as $X\to X_2\to X_1$, where X_2 is the closed image of $X$ by $j$, $h$ the canonical injection and $u$ an open immersion (I, 9.5.3). In other words, one may suppose that X_1 is integral, or also of the form $X_1=\mathrm{Spec}(B)$, where $B$ is an integral and finite $A$-algebra, containing $A$ since $g$ is dominant. If $\mathfrak{n}$ is the prime ideal of $B$ corresponding to the point $x$, the hypothesis that $A$ is unibranch then implies (14.4.1.2) that the morphism $\mathrm{Spec}(B_{\mathfrak{n}})\to \mathrm{Spec}(A)$ is surjective, that is, $\mathrm{Spec}({\mathcal{O}}_{X,x})\to \mathrm{Spec}({\mathcal{O}}_{Y,y})$ is surjective. Q.E.D.

Corollary (14.4.2).

Let $f:X\to Y$ be a morphism locally of finite type, $y$ a geometrically unibranch point of $Y$, $\eta$ the generic point of the unique irreducible component Y_0 of $Y$ containing $y$. The following conditions are equivalent:

a) $f$ is universally open at all points of $X_y$ (or, what comes to the same (14.3.3.1, (ii)), at the maximal points of $X_y$).

b) For every $x\in X_y$, there exists an irreducible component $Z$ of $X$ containing $x$ and equidimensional over $Y$ at the point $x$ (13.2.2).

b') For every $x\in X_y$ and every open neighbourhood $U$ of $x$ in $X$, one has $\dim (U\cap X_{\eta })\ge {\dim }_x(U\cap X_y)$.

b'') For every open $U$ of $X$, one has $\dim (U\cap X_{\eta })\ge \dim (U\cap X_y)$.

When moreover $Y$ is locally Noetherian, these conditions are still equivalent to the following:

c) $f$ is open at all points of $X_y$ (or, what comes to the same (14.3.3.1, (ii)), at the maximal points of $X_y$).

The equivalence of a) and c) when $Y$ is locally Noetherian results from (14.4.1); conditions b) or b'), applied to the maximal points of $X_y$, entail a) by virtue also of (14.3.3.1, (ii)); finally, b') and b'') are equivalent, since

   dim(U ∩ X_y) = sup_x(dim_x(U ∩ X_y)).

It remains to see that condition a) entails b) and b') at every point $x\in X_y$. Set $d={\dim }_x(X_y)$, and let $x^{\prime }$ be the generic point of an irreducible component of $X_y$ containing $x$ and of dimension $d$. By virtue of a) and of (14.4.1), there is an irreducible component $Z$ of $X$ containing $x^{\prime }$ and equidimensional over $Y$ at the point $x^{\prime }$, hence such that ${\dim }_{x^{\prime }}(Z\cap X_y)=\dim (Z\cap X_{\eta })$. But by construction ${\dim }_{x^{\prime }}(Z\cap X_y)={\dim }_{x^{\prime }}(X_y)=d$, and ${\dim }_x(Z\cap X_y)\le \dim (Z\cap X_y)=d$; taking (13.1.6) into account, this proves that $Z$ is equidimensional over $Y$ at the point $x$; hence a) entails b). Moreover, one has dim(X_η) ≥ dim(Z ∩ X_η) = d = dim_x(X_y). Replacing $X$ by an open neighbourhood $U$ of $x$, one sees thus that a) entails b'). Q.E.D.

Corollary (14.4.3).

Let $Y$ be a geometrically unibranch prescheme, $f:X\to Y$ a morphism locally of finite type. The following conditions are equivalent:

a) $f$ is universally open.

b) For every open $U$ of $X$, every $y\in Y$ and every generization $y^{\prime }$ of $y$ one has $\dim (U\cap X_{y^{\prime }})\ge \dim (U\cap X_y)$.

If moreover $Y$ is locally Noetherian, these conditions are also equivalent to the following:

c) $f$ is open.

Corollary (14.4.4) (Chevalley's criterion).

Let $f:X\to Y$ be a morphism locally of finite type.

(i) If $f$ is equidimensional at a point $x\in X$ (13.3.2) and if $y=f(x)$ is a geometrically unibranch point of $Y$, $f$ is universally open at the point $x$.

(ii) If $Y$ is geometrically unibranch, $f$ is universally open at all points of $X$ where $f$ is equidimensional, and the set of these points is open in $X$. In particular, if $f$ is equidimensional, it is universally open.

Assertion (ii) is a trivial consequence of (i), since one already knows that the set of points where $f$ is equidimensional is open (13.3.2). As for assertion (i), it results from the fact that the hypothesis implies that condition b) of (14.4.1) is satisfied at the generic point of an irreducible component of $X_y$ containing $x$, taking (13.3.1) into account; it therefore suffices to apply (14.4.1).

Remark (14.4.5).

One can prove that if $Y$ is locally Noetherian, and if all the generizations of $y=f(x)$ are geometrically unibranch points of $Y$ (cf. (6.15.2)), then, if $f$ is equidimensional at the point $x$, it is universally open in a neighbourhood of $x$.

Corollary (14.4.6).

Let $Y$ be a locally Noetherian prescheme, $f:X\to Y$ a morphism locally of finite type. Let $y$ be a geometrically unibranch point of $Y$, and suppose in addition that for every $x\in f^{-1}(y)$, the ring ${\mathcal{O}}_x$ is equidimensional. Then the following conditions are equivalent:

a) $f$ is equidimensional (13.3.2) at all points of $f^{-1}(y)$.

b) $f$ is open at all points of $f^{-1}(y)$.

c) $f$ is universally open at all points of $f^{-1}(y)$.

Indeed, c) trivially implies b), and a) implies c) by virtue of (14.4.4); finally, in view of the hypotheses on ${\mathcal{O}}_y$ and ${\mathcal{O}}_x$, b) implies a) by (14.2.2). More generally:

Proposition (14.4.7).

Let $Y$ be a locally Noetherian prescheme, $f:X\to Y$ a morphism locally of finite type, $x$ a point of $X$, such that $y=f(x)$ is geometrically unibranch. The following conditions are equivalent:

a) $f$ is equidimensional (13.3.2) at the point $x$.

b) The ring ${\mathcal{O}}_x$ is equidimensional and $f$ is open at the generic points of the irreducible components of $f^{-1}(y)$ containing $x$ (hence also at every point of such a component).