# countable union of closed subschemes over uncountable field

I am looking for a reference for the following well-known fact:

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 $\bigcup Z_i(k) \neq X(k)$.

Thanks!

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You also need $k$ to be algebraically closed (otherwise one could well have $X(k) = \emptyset$). –  ulrich Aug 26 '11 at 9:12
I'm glad you asked the question. You'd think this would discussed in some standard textbook, but I've never seen it. Most complex algebraic geometers use a sledge hammer (Baire category theorem) but it's certainly more elementary than that. –  Donu Arapura Aug 26 '11 at 12:19
It should be possible to give a bare hands proof that given a countable collection of nonzero polynomials $f_i$ in $n$-variables, there exist a point such that $f_i(p)\not=0$ simultaneously. –  Donu Arapura Aug 26 '11 at 12:26
Yes, I left out algebraically closed as a hypothesis. It's frustrating that there (doesn't seem to be) a standard place to quote such a well-known fact. –  Sue Sierra Aug 26 '11 at 18:06
As luck would have it, a related question just came up in my research today: does an algebraic torus over $k$ have a dense cyclic subgroup? If $k$ is uncountable and algebraically closed, the above-mentioned fact shows the answer is yes: just choose a generator outside the (countable) union of all proper Zariski closed subgroups. But if $k$ is the (countable) algebraic closure of a finite field, the answer is no, since every element of $k^\times$ has finite order. –  Michael Thaddeus Aug 28 '11 at 3:59

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 $X$).

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.

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. EDIT 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 $\cup_i Z_i$.

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I do not know a reference, but the following short argument seems to work.

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.

You can find this stated as a hint in an Exercise V.4.15 (c) in Hartshorne.

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As ulrich's correctly comments above, the only place that I did not argue (namely, the base case of the induction) is also the place where there is a missing assumption: the field should be algebraically closed! The argument sketched here requires curves over $k$ to have uncountably many points. –  M P Aug 26 '11 at 9:23