1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.