Why can't subvarieties separate? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:02:18Z http://mathoverflow.net/feeds/question/12892 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12892/why-cant-subvarieties-separate Why can't subvarieties separate? Richard Kent 2010-01-25T03:00:27Z 2010-01-25T04:16:19Z <p>I'm posting my answer to <a href="http://mathoverflow.net/questions/5146/algebraic-geometry-versus-complex-geometry" rel="nofollow">this question</a> as its own question:</p> <p>Let $V$ be an irreducible projective variety over $\mathbb{C}$. Let $U$ be a Zariski open set in $V$. I'll use $V(\mathbb{C})$ and $U(\mathbb{C})$ to mean $V$ and $U$ equipped with their Euclidean topologies, respectively.</p> <p>What is the easiest proof that $U(\mathbb{C})$ is connected?</p> <p>Here's the proof I know: Suppose that $U(\mathbb{C})$ can be written as a disjoint union of two open sets $A$ and $B$. Since the complement of $U$ in $V$ is a variety of smaller dimension than $V$, a theorem of Remmert and Stein implies that the closures $\overline{A}$ and $\overline{B}$ of $A$ and $B$ in $V(\mathbb{C})$ are projective analytic sets. By Chow's theorem that projective analytic sets are algebraic, $\overline{A}$ and $\overline{B}$ are subvarieties of $V$. Since they're proper, $V$ is not irreducible, and we have a contradiction.</p> <p>I guess I'm really asking for the most elementary argument, as I think the above argument is nice intuitively. A reference would be fine.</p> <p>(To avoid going through the same discussion in the comments that happened at the other question, let me point out that I am aware that irreducible varieties are connected and that $U$ is itself a variety in the sense that it is locally affine. It is just not obvious to me that it is irreducible (without appealing to the above argument).)</p> http://mathoverflow.net/questions/12892/why-cant-subvarieties-separate/12896#12896 Answer by Pete L. Clark for Why can't subvarieties separate? Pete L. Clark 2010-01-25T03:34:34Z 2010-01-25T04:16:19Z <p>[This has been completely rewritten at the request of Richard Kent.]</p> <p>Let $X$ be an irreducible topological space and $U$ a non-empty open subset of $X$. Then $U$ is also irreducible -- see e.g. Proposition 141 on page 88 <a href="http://math.uga.edu/~pete/integral.pdf." rel="nofollow">here</a>. (Surely it's also in Hartshorne and lots of other places, but one of the advantages of typing up your own notes is to be able to easily point to a reference because you know exactly where it is.)</p> <p>Thus the question reduces to the fact that if $X_{/\mathbb{C}}$ is an irreducible complex variety, then $X(\mathbb{C})$ with its "Euclidean topology" is connected. For this, see e.g. Section VII.2 of Shafarevich's <em>Basic Algebraic Geometry II</em>. (Again, there are other places, but I think his discussion is especially good.) He gives two different proofs, one of which is a simple induction on the dimension. </p>