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The union of two affine varieties can be expressed as $$ \mathbb{V}(\{F_i\}_{i\in I}) \cup \mathbb{V}(\{F_j\}_{i\in J}) = \mathbb{V}(\{F_iF_j\}_{(i,j) \in I\times J}). $$ We want to generalize this to an infinite union. Let $\{A_i\}_{i \in I}$ be an (infinite) partition of the indexing set $A$; then the left-hand side becomes $$ \bigcup_{i \in I} \mathbb{V}(\{F_\alpha\}_{\alpha \in A_i}), $$ while the defining 'polynomials' on the right hand side must be products of infinitely many polynomials. However, any polynomial must be a finite sum of monomials, so an infinite union of affine varieties is never an affine variety.

Is this correct? If not, are there examples of infinite unions that are varieties, and infinite unions that are not?

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up vote 3 down vote accepted

No. For example, an infinite union of copies of the same variety is always a variety.

You're making the same mistake as in your other question about $\text{U}(n)$: the problem isn't to show that the naive way to make the set a variety fails, but to show that no other way works. In this case, try to show that any affine variety has finitely many irreducible components, so any infinite union of affine varieties with infinitely many irreducible components can't be an affine variety.

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Right, thanks. So is the statement 'any infinite union of distinct affine varieties is not an affine variety' true? (As, if they weren't distinct, we would not necessarily be able to infer that there are infinitely many irreducible components in our union). – Sputnik May 31 '11 at 16:41
Yes, that's right. – Qiaochu Yuan May 31 '11 at 16:43
Even a finite union of affine varieties is not necessarily affine, for instance projective space $\mathbb P^n= \bigcup_{i=0}^n U_i$ where $U_i \cong \mathbb A^n$. In fact, any positive dimensional projective variety is not affine and has a finite covering by open affine subsets. – Parsa Apr 27 '12 at 8:05
@Parsa: here by "union" I think the OP implicitly means "union in $\mathbb{A}^n$." – Qiaochu Yuan Apr 27 '12 at 8:06

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