I would like to get a reference of the following fact.

Let $A\subseteq B$ be affine domains over an algebraically closed field of characteristic zero. If $Q(A)$ is algebraically closed in $Q(B)$, show that any genral fiber of the associated morphism of schemes is irreducible, or in other words, there exists a non-empty open set $V$ in Max $A$, such that for any maximal ideal $\mathfrak{m}\in$ Max $A$, the extended ideal $\mathfrak{m}B$ is irreducible.

Thank you in advance.

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It seems to me that what you may be looking for is [EGA IV$_3$, 9.7.8], which says in particular that if $S$ is an irreducible scheme with function field $K$ and $X$ is an $S$-scheme of finite presentation such that $X_K$ is geometrically irreducible, then there is a nonempty open $U \subset S$ such that for every $s \in U$ the fiber $X_{k(s)}$ is geometrically irreducible.

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  • $\begingroup$ I Know that $X_K$ is irreducible. But don't know if it is geometrically irreducible, actually that is the problem. $\endgroup$ – sagnik chakraborty Apr 24 '15 at 16:39
  • $\begingroup$ @sagnikchakraborty: But you are assuming that $Q(A)$ is algebraically closed in $Q(B)$, so [EGA IV$_2$, 4.5.9] applies: an irreducible $K$-scheme $X_K$ is geometrically irreducible if and only if its function field is a primary extension of $K$. $\endgroup$ – Kestutis Cesnavicius Apr 24 '15 at 18:41
  • $\begingroup$ But why is $Q(A)$ algebraically closed in the field of fractions of $B\otimes_AQ(A)$? $\endgroup$ – sagnik chakraborty Apr 24 '15 at 19:05
  • $\begingroup$ Because the field of fractions of $B \otimes_A Q(A)$ is the same as the field of fractions of $B$ (localization is transitive). $\endgroup$ – Kestutis Cesnavicius Apr 24 '15 at 19:14
  • $\begingroup$ Sorry, my mistake. It's $B\otimes_kQ(A)$, and not $B\otimes_AQ(A)$, where $A$ and $B$ were $k$-affine domains to start with. $\endgroup$ – sagnik chakraborty Apr 24 '15 at 19:18

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