Loading web-font TeX/Math/Italic

The Stacks project

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

\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 37.27.7 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


Comments (0)


Your email address will not be published.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.