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From: Omega John on 16 Feb 2010 16:47
Hi

True or false?
From: Dan Cass on 16 Feb 2010 22:39
> Hi
>
> True or false?

The question was in the title of the post, namely
--> Disjoint union of two algebraic varieties is an alg. variety?

If by disjoint union you mean the union of two sets
which happen to be disjoint, then it would seem yes.
For example the disjoint circles
A... x^2 + y^2 = 9
B... x^2 + y^2 = 16
are both "varieties" in the sense of being the zero set
of polynomials.
A: x^2 + y^2 - 9 = 0
B: x^2 + y^2 - 16 = 0.

In this case the union of A and B is another variety,
since it is the zero set of
(x^2 + y^2 - 9)*(x^2 + y^2 - 16).
From: Hagen on 17 Feb 2010 17:33
> > Hi
> >
> > True or false?
>
> The question was in the title of the post, namely
> --> Disjoint union of two algebraic varieties is an
> alg. variety?
>
> If by disjoint union you mean the union of two sets
> which happen to be disjoint, then it would seem yes.
> For example the disjoint circles
> A... x^2 + y^2 = 9
> B... x^2 + y^2 = 16
> are both "varieties" in the sense of being the zero
> set
> of polynomials.
> A: x^2 + y^2 - 9 = 0
> B: x^2 + y^2 - 16 = 0.
>
> In this case the union of A and B is another
> variety,
> since it is the zero set of
> (x^2 + y^2 - 9)*(x^2 + y^2 - 16).

A variety by definition must be irreducible as a
topological space with respect to the Zariski topology.
Your example demonstrates that the union of the
two circles is NOT a variety in the obvious sense.

The problem with the questions as stated is this:

If yout take two varieties given somehow, then the
disjoint union apriori is just a set. To turn it into
an object that can be considered in algebraic geometry
you have introduce a structure sheaf - in particular a
topology - on it.

H
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