Quoting Chris: X = A^2 - (0,0) is not isomorphic to an affine variety. To show this I'm supposed to work out k[X]. I don't know how to do this.

Also, I'm unclear on how to break varieties up into irreducible components. For example,

Take two of the three defining equations for the projective twisted cubic: J= X = Zer(J) should be reducible. How do you seperate it into its irr. components?

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Quoting Patrick: 1) I(X) = {f in k[x,y] : f(0,0)=0} = (=notation for x*k[x,y]+y*k[x,y]). k[X]=k[x,y]/I(X)= k. (but I don't know why you need this). To show X is not an affine variety, you need only show that Z(I(X)) is not equal to X. This is clear since Z(I(X))=Z()={(a,b) in k^2 : a=0 or b=0} not= (0,0).

2) Z(J1+J2+J3)=Z(J1) intersect Z(J2) intersect Z(J3). So if Ji are all prime, then Z(Ji) are all irred (by Nullstellensatz - very important). The polynomials you gave below are prime, so that's the most natural way to look at Z(J). But this isn't an irred decomp. For that you need unions... The only "natural" thing I can think of in that case (and we have to look at natural things in alg geo if we don't want to go crazy with wicked computations) is if J=J1*J2*J3 where each of Ji are prime (product instead of sum). Then Z(J)=Z(J1) union Z(J2) union Z(J3) is an irred decomp. Thus a prime factorization of an ideal naturally corresponds to an irred decomp.

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Thnaks Patrick, in (1), we need to show X is not isomorphic to an affine variety (this is Shaferevich's definition of an affine quasiprojective variety), and that is why we need to look at k[X]. For example A - 0 = Y is not an affine closed set, but it is isomorphic to a hyperbola which is. Thus it is considered an affine variety.

I'll look more at (2) to see if I can find the ideals I want. I'll give you the solutions to post when I have them all nice and pretty.

Lates,

Chris

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So K[X] is a domain iff I(X) is prime iff X is irred. But in our example, we have X=A^2-0, so I(X)= which is prime, and K[X]=K is a domain - so does that tell us that X is irred?

Patrick