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Let Y be locally Noetherian, and consider a projective morphism $f:X \rightarrow Y$ such that the map $\textbf{Spec} f_\ast \mathcal{O}_X \rightarrow Y $ is universally injective. Let $C \rightarrow Y$ be a morphism of schemes with $C$ connected. How could I show that $X \times_Y C$ is connected?

So far, I have been trying to show that the ring of global sections in the fiber product doesn't have any non-trivial idempotents by using that C doesn't have any. I have also noted that the map from the fiber product to C is projective, but no luck there. Thankful, Heidar

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  • Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. – Kasper Apr 26 '13 at 13:42
  • Thank you! I will edit it a bit. – Heidar Svan Apr 26 '13 at 14:15
  • QiL: Fixed the statement now. If you see how to do this, please let me know. – Heidar Svan Apr 29 '13 at 10:44
  • Using the new hypothesis, prove that each fiber of $X\to Y$ is geometrically connected, hence each fiber of $f_C: X\times_Y C\to C$ is connected. Then use the properness of $f_C$ and the connectedness of $C$ to show $X\times_Y C$ is connected. –  Apr 30 '13 at 02:32
  • QiL: I don't see how to use that $Spec f_* \mathcal{O}_X \rightarrow Y$ finite, surjective implies connected fibers. Any further hints? (sorry if I am missing something obvious). – Heidar Svan Apr 30 '13 at 11:00
  • @HeidarSvan: I am very sorry, I confused the morphism with the map $O_Y\to f_O_X$. Your initial statement is correct. When $\mathrm{Spec}(f_O_X)\to Y$ is universally injective, it is purely inseparable (see "Algebraic geometry and arithmetic curves", 5.3.13). For any $y\in Y$, $k(y)\to H^0(X_y, O_{X_y})$ is then purely inseparable (op. cit., 5.3.14). So, when you extend to an algebraic closure $\bar{k}$ of $k(y)$, the global sections of the structure sheaf of $X_{\bar{k}}$ is still a local algebra, hence $X_{\bar{k}}$ is connected. –  May 03 '13 at 04:40
  • Heidar Svan: write @QiL'8 instead of QiL in your comment if you want me be aware of your comment. –  May 03 '13 at 04:41
  • One can also use Stein factorization : $X\to {\bf Spec}(f_*O_X)\to Y$. It is known that the first map has geometrically connected fibers, and the second has geometrically connected fibers by the universally injective hypothesis. –  May 03 '13 at 05:34

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