I'm posting my answer to this question as its own question:

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.

What is the easiest proof that $U(\mathbb{C})$ is connected?

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.

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.

(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).)

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    $\begingroup$ It seems to me that your question is equivalent to asking why a variety over the complex numbers is connected in the Zariski topology iff its complex points are connected in the analytic topology: am I missing a nuance here? For that, Shafarevich's Basic Algebraic Geometry has a proof, and in general proofs in this book tend to be both intuitive and elementary (albeit sometimes with details left out). $\endgroup$ Jan 25, 2010 at 3:23
  • $\begingroup$ I don't think that's the same question. The intuition you get from thinking about manifolds is supposed to be that subvarieties have real codimension two and so shouldn't separate. The issue is that the singular locus of $V$ could separate. $\endgroup$ Jan 25, 2010 at 3:33
  • $\begingroup$ (Also, you can be reducible and connected in the analytic topology, so that direction doesn't go.) $\endgroup$ Jan 25, 2010 at 3:35
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    $\begingroup$ @Richard: a Zariski-open subset of an irreducible topological space is irreducible (or empty!), so I think my answer below does apply. $\endgroup$ Jan 25, 2010 at 3:38

1 Answer 1


[This has been completely rewritten at the request of Autumn Kent.]

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 here. (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.)

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 Basic Algebraic Geometry II. (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.

  • $\begingroup$ As I mentioned in the question, I am aware of this. The real question is why is the smooth locus connected in the analytic topology. $\endgroup$ Jan 25, 2010 at 3:37
  • $\begingroup$ Sorry, I'm afraid I'm still not understanding you. The smooth locus is Zariski open, so it's irreducible, so it's analytically connected by the theorem referred to above. $\endgroup$ Jan 25, 2010 at 3:41
  • $\begingroup$ I'm sorry. I thought it was clear in the question. I do not see why a Zariski open set in an irreducible variety is irreducible. $\endgroup$ Jan 25, 2010 at 3:43
  • $\begingroup$ This is actually a matter of general topology. See e.g. Proposition 141 on page 88 of math.uga.edu/~pete/integral.pdf. It relies on the previous Proposition, whose proof is left as an exercise, but I believe you will find it to be straightforward. $\endgroup$ Jan 25, 2010 at 3:53
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    $\begingroup$ In Griffiths&Harris, page 21, it is proved that a variety is irreducible if and only if its smooth locus is connected. $\endgroup$ Mar 29, 2010 at 21:17

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