Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This should be an elementary question for anyone who knows SGA by heart (alas, not for me). It smells a lot like a descent problem. All schemes are supposed to be noetherian, and all morphisms to be locally of finite presentation.

Let $X$ be a scheme of finite type over a (base) scheme $S$, and let $R \subseteq X(S)$ be a subset of the set of $S$--rational points of $X$. Denote by $\overline R$ the smallest closed subscheme of $X$ whose $S$--rational points contain $R$.

Let $f:S'\to S$ be a (base--change) morphism of schemes, write $X' := X\times_SS'$ and denote by $R'$ the image of $R$ in $X(S') = X'(S')$. Again, let $\overline{R'}$ be the smallest closed subscheme of $X'$ whose $S'$--rational points contain $R'$. Then, $\overline{R'}$ is contained in $\overline R\times_SS'$, and the question is:

Suppose $f:S'\to S$ is flat. Does the equality $\overline{R'} = \overline R \times_SS'$ hold?

Clearly some hypothesis on $f$ is needed, and I just guess it's flatness.

share|cite|improve this question

1 Answer 1

up vote 5 down vote accepted

It seems that the condition you need is that the generic points of $S'$ go to generic points of $S$; this is much weaker than flatness.

Assuming this condition, we can reduce to the case that $S$ and $S'$ are both $Spec$ of some field and we can also assume that $\overline{R} = X$. If $X$ is a variety over a field $k$ any Zariski dense subset of $X(k)$ is also dense as a subset of $X(K)$ where $K$ is any extension field of $X$, proving the claim.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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