The Stacks project

Lemma 37.33.7. Let $f : X \to S$ be a morphism of affine schemes, which is of finite presentation with geometrically irreducible fibres. Then there exists a diagram as in Lemma 37.33.1 such that in addition $f_0$ has geometrically irreducible fibres.

Proof. Apply Lemma 37.33.1 to get a cartesian diagram

\[ \xymatrix{ X_0 \ar[d]_{f_0} & X \ar[l]^ g \ar[d]^ f \\ S_0 & S \ar[l]_ h } \]

of affine schemes with $X_0 \to S_0$ a finite type morphism of schemes of finite type over $\mathbf{Z}$. By Lemma 37.26.7 the set $E \subset S_0$ of points where the fibre of $f_0$ is geometrically irreducible is a constructible subset. By Lemma 37.26.2 we have $h(S) \subset E$. Write $S_0 = \mathop{\mathrm{Spec}}(A_0)$ and $S = \mathop{\mathrm{Spec}}(A)$. Write $A = \mathop{\mathrm{colim}}\nolimits _ i A_ i$ as a direct colimit of finite type $A_0$-algebras. By Limits, Lemma 32.4.10 we see that $\mathop{\mathrm{Spec}}(A_ i) \to S_0$ has image contained in $E$ for some $i$. After replacing $S_0$ by $\mathop{\mathrm{Spec}}(A_ i)$ and $X_0$ by $X_0 \times _{S_0} \mathop{\mathrm{Spec}}(A_ i)$ we see that all fibres of $f_0$ are geometrically irreducible. $\square$


Comments (0)


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Your email address will not be published.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 05FH. Beware of the difference between the letter 'O' and the digit '0'.