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Hartshorne wrote in his book's Appendix B that it can be easily proved that a complex algebraic variety is connected in the usual topology if and only if it is connected in Zariski topology. How can it be proved?

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By the way, the term "the usual topology" may be confusing(some may think Zariski topology is the usual one). Is there a better term? – Makoto Kato Nov 22 '12 at 23:37
You should write "scheme of finite type over $\mathbb C$" instead of "algebraic variety", since in Hartshorne "variety" assumes "integral scheme" which implies irreducible, hence connected. You could say "analytic topology" I guess, no one could possibly confuse this with the Zariski topology. But hey, this post is quite old anyway. – Patrick Da Silva Mar 23 at 12:20
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Fredrik already explained why the connectedness for the complex topology implies the connectedness for the Zariski topology.

The converse is harder. In the projective case, see Georges's wonderful answer. Let me give a reference for the general case. Let $X_1, \dots, X_n$ be the (Zariski) irreducible components of $X$. By Shafarevich, Basic Algebraic Geometry 2, Chapter VII, § 2.2, Theorem 1, $X_i(\mathbb C)$ is connected in complex topology for all $i\le n$. Let $Y$ be a connected component of $X(\mathbb C)$. Then $Y$ is the union of some $X_i(\mathbb C)$ and $X(\mathbb C)\setminus Y$ is the union of the other $X_i(\mathbb C)$'s. So $Y$ and $X\setminus Y$ are both Zariski closed and disjoint. This implies that $X$ is not Zariski connected if $X(\mathbb C)$ is not connected for the complex topology.

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If $X$ is a complex projective algebraic variety, the canonical morphism $$ H^i(X, F)\stackrel {\cong}{\to} H^i(X^{an},F^{an})$$ is an isomorphism of complex finite-dimensional vector spaces for all coherent sheaves on the algebraic variety $X$.
In particular for $i=0$ and $F=\mathcal O_X$ we get an isomorphism $$\Gamma(X, \mathcal O_X) \stackrel {\cong}{\to} \Gamma(X^{an}, \mathcal O_X^{an}) \quad (\bigstar)$$ Now $X$ is connected in the Zariski topology iff the left-hand side has dimension $1$, and similarly $X^{an}$ is connected in the classical topology iff the right-hand side has dimension $1$.
The isomorphism $(\bigstar)$ then implies that$X$ is connected in its Zariski topology iff $X^{an}$ is connected in its classical topology.

The proof only works for projective (or slightly more generally for complete) varieties and uses the full power of GAGA but , hey, who can resist the pleasure of shooting at flies with ballistic missiles?

As an answer to Makoto's comment, in order to prove that on a connected reduced compact analytic space all holomorphic functions $f:X\to \mathbb C$ are constant, it may be amusing in the spirit of nuking mosquitoes to invoke Remmert's theorem according to which $f(X)$ is analytic in $\mathbb C$ , since $f$ is proper. Hence it is a point since it is compact and connected!

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"$X^{an}$ is connected in the classical topology iff the right-hand side has dimension $1$". If $X$ is smooth, I can see this using the maximal absolute value principle of an analytic function. Could you explain this in general case? – Makoto Kato Nov 22 '12 at 13:40
Dear Makoto, this follows from Theorem (5.9) in Chapter II of [Demailly's notes ](…) , page 104 (he supposes the variety irrreducible, but that is not necessary, as long as it is connected.) Or see my edit, if you want a mischievous answer... – Georges Elencwajg Nov 22 '12 at 14:40
Dear @Georges, I wish I could vote your answers up more than once. – Keenan Kidwell Nov 22 '12 at 16:29
Dear @Keenan, I really appreciate your kind words: thank you very much. – Georges Elencwajg Nov 22 '12 at 18:14
Dear Georges, Thanks for the link.Perhaps you mean Corollary (5.8) in Chapter II of the Demailly's book which states if $X$ is a compact irreducible analytic space, then every holomorphic function on $X$ is constant? – Makoto Kato Nov 22 '12 at 19:32

It is easy at least in the affine case. (so the variety $X=\text{Spec }\mathbb{C}[x_1,\cdots,x_n]$ for some $n$.

If $X$ is disconnected (say, in two components for simplicity) in the usual topology, then its global section decomposes into direct products: $\Gamma(X,\mathcal{O}_X)= \Gamma(U,\mathcal{O}_X) \times \Gamma(V,\mathcal{O}_X)$. Taking $\text{Spec }$ of this gives $X$ back, and it is a standard result (Hartshorne, 2.19, Chapter II, for example), that $\text{Spec } A$ is disconnected if and only if $A$ can be written as a direct product of nonzero rings. Much of this part is wrong. See Zhen Lin's comment below.

On the other hand, if assume $X$ is disconnected in the Zariski topology. Since the Zariski topology is contained in the usual topology, every clopen (closed and open) set is also clopen in the usual topology. A topological space is disconnected if and only if there exists a non-trivial clopen set.

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$U$ is not necessarily Zariski open. What does $\Gamma(U,\mathcal{O}_X)$ mean? – Makoto Kato Nov 22 '12 at 11:58
Can you explain? If a topological space is disconnected, then each component is open. (by $\Gamma(U,\mathcal{O_X})$ I mean the structure sheaf evaluated at $U$. Often written $\mathcal{O}_X(U)$) – Fredrik Meyer Nov 22 '12 at 12:03
You can define $\mathscr{O}_X$ as the sheaf of analytic functions on $X$, when $X$ is an analytic space. Then it makes sense to talk about $\Gamma (U, \mathscr{O}_X)$ when $U$ is an open subset in the metric topology. But it's not immediately clear that $\mathscr{O}_X$ on a Zariski-open set $U$ agrees with the set of regular functions on $U$... – Zhen Lin Nov 22 '12 at 12:04
@ZhenLin: I see. I was naïvely assuming that a complex algebraic variety just ment that the regular functions were the same. – Fredrik Meyer Nov 22 '12 at 12:13
Dear @Zhen, not only is it not clear that the set of regular functions on an affine complex variety coincides with the set of holomorphic functions, but it is completely false: there are many entire functions on $\mathbb C$ which are not polynomials ( the exponential function comes to mind ...) – Georges Elencwajg Nov 22 '12 at 12:59

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