The Stacks project

Lemma 37.26.4. Let $f : X \to Y$ be a morphism of schemes. Assume

  1. $Y$ is irreducible with generic point $\eta $,

  2. $X_\eta $ is geometrically connected, and

  3. $f$ is of finite type.

Then there exists a nonempty open subscheme $V \subset Y$ such that $X_ V \to V$ has geometrically connected fibres.

Proof. Choose a diagram

\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X_ V \ar[r] \ar[d] & X \ar[d]^ f \\ Y' \ar[r]^ g & V \ar[r] & Y } \]

as in Lemma 37.22.8. Note that the generic fibre of $f'$ is geometrically connected (for example by Lemma 37.26.3). Suppose that the lemma holds for the morphism $f'$. This means that there exists a nonempty open $W \subset Y'$ such that every fibre of $X' \to Y'$ over $W$ is geometrically connected. Then, as $g$ is an open morphism by Morphisms, Lemma 29.36.13 all the fibres of $f$ at points of the nonempty open $V = g(W)$ are geometrically connected, see Lemma 37.26.3. In this way we see that we may assume that the irreducible components of the generic fibre $X_\eta $ are geometrically irreducible.

Let $Y'$ be the reduction of $Y$, and set $X' = Y' \times _ Y X$. Then it suffices to prove the lemma for the morphism $X' \to Y'$ (for example by Lemma 37.26.3 once again). Since the generic fibre of $X' \to Y'$ is the same as the generic fibre of $X \to Y$ we see that we may assume that $Y$ is irreducible and reduced (i.e., integral, see Properties, Lemma 28.3.4) and that the irreducible components of the generic fibre $X_\eta $ are geometrically irreducible.

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. 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 37.22.5. Let $Z_{i, j} = X_ i \cap X_ j$. Let

\[ \{ 1, \ldots , n\} \times \{ 1, \ldots , n\} = I \amalg J \]

where $(i, j) \in I$ if $Z_{i, j, \eta } = \emptyset $ and $(i, j) \in J$ if $Z_{i, j, \eta } \not= \emptyset $. After shrinking $Y$ we may assume that $Z_{i, j} = \emptyset $ for all $(i, j) \in I$, see Lemma 37.22.1. After shrinking $Y$ we obtain that $X_{i, y}$ is geometrically irreducible for each $i$ and all $y \in Y$, see Lemma 37.25.5. After shrinking $Y$ some more we achieve the situation where each $Z_{i, j} \to Y$ is flat and of finite presentation for all $(i, j) \in J$, see Morphisms, Proposition 29.27.1. This means that $f(Z_{i, j}) \subset Y$ is open, see Morphisms, Lemma 29.25.10. We claim that

\[ V = \bigcap \nolimits _{(i, j) \in J} f(Z_{i, j}) \]

works, i.e., that $X_ y$ is geometrically connected for each $y \in V$. Namely, the fact that $X_\eta $ is connected implies that the equivalence relation generated by the pairs in $J$ has only one equivalence class. Now if $y \in V$ and $K \supset \kappa (y)$ is a separably closed extension, then the irreducible components of $(X_ y)_ K$ are the fibres $(X_{i, y})_ K$. Moreover, we see by construction and $y \in V$ that $(X_{i, y})_ K$ meets $(X_{j, y})_ K$ if and only if $(i, j) \in J$. Hence the remark on equivalence classes shows that $(X_ y)_ K$ is connected and we win. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 37.26: Connected components of fibres

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 055G. Beware of the difference between the letter 'O' and the digit '0'.