Proposition 41.6.1. Sections of unramified morphisms.
Any section of an unramified morphism is an open immersion.
Any section of a separated morphism is a closed immersion.
Any section of an unramified separated morphism is open and closed.
41.6 Topological properties of unramified morphisms
The first topological result that will be of utility to us is one which says that unramified and separated morphisms have “nice” sections. The material in this section does not require any Noetherian hypotheses.
Proof. Fix a base scheme $S$. If $f : X' \to X$ is any $S$-morphism, then the graph $\Gamma _ f : X' \to X' \times _ S X$ is obtained as the base change of the diagonal $\Delta _{X/S} : X \to X \times _ S X$ via the projection $X' \times _ S X \to X \times _ S X$. If $g : X \to S$ is separated (resp. unramified) then the diagonal is a closed immersion (resp. open immersion) by Schemes, Definition 26.21.3 (resp. Morphisms, Lemma 29.35.13). Hence so is the graph as a base change (by Schemes, Lemma 26.18.2). In the special case $X' = S$, we obtain (1), resp. (2). Part (3) follows on combining (1) and (2). $\square$
We can now explicitly describe the sections of unramified morphisms.
Theorem 41.6.2. Let $Y$ be a connected scheme. Let $f : X \to Y$ be unramified and separated. Every section of $f$ is an isomorphism onto a connected component. There exists a bijective correspondence In particular, given $x \in X$ there is at most one section passing through $x$.
\[ \text{sections of }f \leftrightarrow \left\{ \begin{matrix} \text{connected components }X'\text{ of }X\text{ such that}
\\ \text{the induced map }X' \to Y\text{ is an isomorphism}
\end{matrix} \right\} \]
Proof. Direct from Proposition 41.6.1 part (3). $\square$
The preceding theorem gives us some idea of the “rigidity” of unramified morphisms. Further indication is provided by the following proposition which, besides being intrinsically interesting, is also useful in the theory of the algebraic fundamental group (see [Exposé V, SGA1]). See also the more general Morphisms, Lemma 29.35.17.
Proposition 41.6.3. Let $S$ is be a scheme. Let $\pi : X \to S$ be unramified and separated. Let $Y$ be an $S$-scheme and $y \in Y$ a point. Let $f, g : Y \to X$ be two $S$-morphisms. Assume
$Y$ is connected
$x = f(y) = g(y)$, and
the induced maps $f^\sharp , g^\sharp : \kappa (x) \to \kappa (y)$ on residue fields are equal.
Then $f = g$.
Proof. The maps $f, g : Y \to X$ define maps $f', g' : Y \to X_ Y = Y \times _ S X$ which are sections of the structure map $X_ Y \to Y$. Note that $f = g$ if and only if $f' = g'$. The structure map $X_ Y \to Y$ is the base change of $\pi $ and hence unramified and separated also (see Morphisms, Lemmas 29.35.5 and Schemes, Lemma 26.21.12). Thus according to Theorem 41.6.2 it suffices to prove that $f'$ and $g'$ pass through the same point of $X_ Y$. And this is exactly what the hypotheses (2) and (3) guarantee, namely $f'(y) = g'(y) \in X_ Y$. $\square$
Lemma 41.6.4. Let $S$ be a Noetherian scheme. Let $X \to S$ be a quasi-compact unramified morphism. Let $Y \to S$ be a morphism with $Y$ Noetherian. Then $\mathop{\mathrm{Mor}}\nolimits _ S(Y, X)$ is a finite set.
Proof. Assume first $X \to S$ is separated (which is often the case in practice). Since $Y$ is Noetherian it has finitely many connected components. Thus we may assume $Y$ is connected. Choose a point $y \in Y$ with image $s \in S$. Since $X \to S$ is unramified and quasi-compact then fibre $X_ s$ is finite, say $X_ s = \{ x_1, \ldots , x_ n\} $ and $\kappa (x_ i)/\kappa (s)$ is a finite field extension. See Morphisms, Lemma 29.35.10, 29.20.5, and 29.20.10. For each $i$ there are at most finitely many $\kappa (s)$-algebra maps $\kappa (x_ i) \to \kappa (y)$ (by elementary field theory). Thus $\mathop{\mathrm{Mor}}\nolimits _ S(Y, X)$ is finite by Proposition 41.6.3.
General case. There exists a nonempty open $U \subset S$ such that $X_ U \to U$ is finite (in particular separated), see Morphisms, Lemma 29.51.1 (the lemma applies since we've already seen above that a quasi-compact unramified morphism is quasi-finite and since $X \to S$ is quasi-separated by Morphisms, Lemma 29.15.7). Let $Z \subset S$ be the reduced closed subscheme supported on the complement of $U$. By Noetherian induction, we see that $\mathop{\mathrm{Mor}}\nolimits _ Z(Y_ Z, X_ Z)$ is finite (details omitted). By the result of the first paragraph the set $\mathop{\mathrm{Mor}}\nolimits _ U(Y_ U, X_ U)$ is finite. Thus it suffices to show that
\[ \mathop{\mathrm{Mor}}\nolimits _ S(Y, X) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _ Z(Y_ Z, X_ Z) \times \mathop{\mathrm{Mor}}\nolimits _ U(Y_ U, X_ U) \]
is injective. This follows from the fact that the set of points where two morphisms $a, b : Y \to X$ agree is open in $Y$, due to the fact that $\Delta : X \to X \times _ S X$ is open, see Morphisms, Lemma 29.35.13. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)