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Just out of curiosity:

Why is it that all polynomials vanishing on an irreducible component in some affine space form a vector subspace?

Thanks for your time.

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Let $\mathbb{A}_k^n$ be affine $n$-space over some algebraically closed field $k$, and let $S\subseteq \mathbb{A}_k^n$ be any subset. Given $f,g\in k[x_1,\ldots,x_n]$, if $f(p)=0$ and $g(p)=0$ for all $p\in S$, then for any $a,b\in k$, $$(af+bg)(p)=af(p)+bg(p)=a\cdot0+b\cdot0=0.$$ Thus $I(S)$, the set of polynomials in $k[x_1,\ldots,x_n]$ which vanish on $S$, forms a vector subspace of $k[x_1,\ldots,x_n]$. In fact, it forms an ideal, because for any $h\in k[x_1,\ldots,x_n]$, $$(hf)(p)=h(p)\cdot f(p)=h(p)\cdot 0=0.$$ When $S$ is an irreducible algebraic set in $\mathbb{A}_k^n$ (a.k.a. an algebraic variety), the ideal $I(S)$ is a prime ideal, and conversely as well.

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Thanks Zev! So by what you did above, an irreducible component of scheme cannot be the union of finitely many proper subspaces. –  math-visitor May 19 '12 at 21:00
    
I don't see how that's related. It is true that an irreducible scheme (in fact, disregarding the scheme structure, any irreducible topological space) cannot be the union of finitely many proper closed subspaces, which follows directly from the definition of "irreducible". But how does that follow from what I've written? –  Zev Chonoles May 19 '12 at 21:04
    
Well, you just proved that the functions vanishing on an irreducible variety form a vector space, and a vector space cannot be the union of finitely many proper subspaces. Do I need to add that we're working over the complexes? –  math-visitor May 19 '12 at 21:13
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