Let $f:X \to Y$ be a flat proper morphism of noetherian schemes of finite type over a field. Assume that $Y$ is an integral scheme and the generic fiber of $f$ is irreducible of dimension $1$. Is it then true that every closed fiber of $f$ is connected?

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    $\begingroup$ Yes (you don't need dimension 1, only that the generic fiber is connected). This is EGA IV, Cor. 15.5.4. $\endgroup$ – abx Mar 25 '15 at 13:15
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    $\begingroup$ I agree with abx's comment if by "connected" you mean "geometrically connected". $\endgroup$ – Laurent Moret-Bailly Mar 25 '15 at 13:19
  • $\begingroup$ @abx, Moret-Bailly: Thank you very much. $\endgroup$ – user45397 Mar 25 '15 at 13:20
  • $\begingroup$ If the field mentioned in the question is algebraically closed, then can we say that the generic fiber of $f$ is geometrically connected (assuming it is irreducible)? If not, is there any known additional assumption when this holds true? $\endgroup$ – user45397 Mar 26 '15 at 12:34
  • $\begingroup$ @user45397 — geometrically connected means connected over $\bar{k}$. Since irreducible implies connected, you are fine. $\endgroup$ – jmc Mar 26 '15 at 12:44

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