countable union of closed subschemes over uncountable field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:02:33Z http://mathoverflow.net/feeds/question/73743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73743/countable-union-of-closed-subschemes-over-uncountable-field countable union of closed subschemes over uncountable field Sue Sierra 2011-08-26T08:22:48Z 2011-08-28T17:32:18Z <p>I am looking for a reference for the following well-known fact:</p> <p>Let $k$ be an uncountable field, and let $X$ be a $k$-variety. Let $Z_1, Z_2, \dots \subseteq X$ be proper closed subschemes. Then $\bigcup Z_i(k) \neq X(k)$.</p> <p>Thanks!</p> http://mathoverflow.net/questions/73743/countable-union-of-closed-subschemes-over-uncountable-field/73750#73750 Answer by MP for countable union of closed subschemes over uncountable field MP 2011-08-26T09:09:48Z 2011-08-26T09:09:48Z <p>I do not know a reference, but the following short argument seems to work.</p> <p>Assume that the dimension of $X$ is at least 1! Argue by induction on the dimension of $X$. Reduce to the case in which the subschemes are irreducible of codimension one. Shrinking $X$ if necessary, reduce also to the case in which $X$ is quasiprojective. Let $L$ be a pencil of integral divisors on $X$. Since the ground-field is uncountable, the pencil $L$ contains uncountably many elements that are integral; let $D \in L$ be an integral element of $L$ that is different from each of the subschemes you want to avoid. By the inductive hypothesis, $D$ contains a point that is not contained in any of the subschemes and you are done.</p> <p>You can find this stated as a hint in an Exercise V.4.15 (c) in Hartshorne.</p> http://mathoverflow.net/questions/73743/countable-union-of-closed-subschemes-over-uncountable-field/73868#73868 Answer by Qing Liu for countable union of closed subschemes over uncountable field Qing Liu 2011-08-27T21:35:18Z 2011-08-28T17:32:18Z <p>Suppose $\dim X>0$ and $k$ is algebraically closed and uncountable. Moreover, if a "variety" is not necessarily irreducible, the $Z_i$ are supposed to have positive codimension in $X$ (otherwise one could take the irreducible components of $X$). </p> <p>As in MP's answer, one can suppose $X$ is affine. By Noether's Normalization Lemma, there exists a finite surjective morphism $p: X\to \mathbb A^m_k$ with $m=\dim X$. Let $Y_i=p(Z_i)$. This is a closed subset of $\mathbb A^m_k$ of positive codimension. Moreover $\mathbb A^m_k(k)=\cup Y_i(k)$ because $k$ is algebraically closed (which implies that $Y_i(k)=p(Z_i(k))$). As $k$ is uncountable, there exists a hyperplane $H$ in $\mathbb A^m$ not contained in any $Y_i$ (note that $H\subseteq Y_i$ is equivalent to $H=Y_i$). So by induction on $m$ we are reduced to the case $m=1$, and the assertion is obvious. </p> <p>Without the hypothesis $k$ algebraically closed, one can show similarly that $X\ne \cup_i Z_i$. This is Exercise 2.5.10 in my book. <b>EDIT</b> In fact this statement is trivial because the generic points of $X$ don't belong to any of the $Z_i$'s. But the proof shows that the set of closed points of $X$ is not contained in $\cup_i Z_i$.</p>