Lemma 37.34.7. 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 37.34.1 such that in addition $f_0$ has geometrically irreducible fibres.
Proof. Apply Lemma 37.34.1 to get a cartesian diagram
of affine schemes with $X_0 \to S_0$ a finite type morphism of schemes of finite type over $\mathbf{Z}$. By Lemma 37.27.7 the set $E \subset S_0$ of points where the fibre of $f_0$ is geometrically irreducible is a constructible subset. By Lemma 37.27.2 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. $\square$
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Your email address will not be published.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)