Give an example of a prevariety that is not a variety
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Answer:
The reason why OP,X=OP,A2OP,X=OP,A2 is because for any point PP in an affine variety YY and U⊆YU⊆Y open, we have that OP,U≅OP,YOP,U≅OP,Y. This should essentially be immediate (what is the definition of the local ring at a point?) The second equality comes from Theorem 1.3.2 (c) of Hartshorne.
That k[x,y]k[x,y] is a UFD is a standard fact in basic abstract algebra. More generally for any UFD RR, the polynomial ring in nn number of variables over RR is also a UFD. For your second question, suppose I take a quotient of two polynomials f=g/hf=g/h with hh and gg coprime. Suppose that there is another representation g′/h′g′/h′ of ff with g′g′ and h′h′ coprime. Then it will follow that h′g=g′hh′g=g′h which means to say that h|h′gh|h′g. By unique factorization, it follows that every prime factor of hh divides h′h′ and in fact h|h′h|h′ since hh divides h′gh′g but not gg.
Similarly we conclude that h′|hh′|h and so upto multiplication by a scalar, h′=hh′=h. Mutadis mutandis the same argument also shows that g′=gg′=g and so the representation of ff as a quotient of two polynomials is unique upto multiplication by a scalar. Thus whenever we choose an element ff of k(x,y)k(x,y), essentially to ask questions about whether a denominator vanishes, it is enough to work with just any representative of ff as a quotient of two polynomials with no common factor.
Thus if f∈k[x,y]mPf∈k[x,y]mP then ff can be written as the quotient of two polynomials g/hg/h with h(P)≠0h(P)≠0 (definition of the localization at mPmP). Conversely if f=g/hf=g/h with h(P)≠0h(P)≠0, then for any other representative g′/h′g′/h′ of ff, h′(P)≠0h′(P)≠0 too for h′h′ is a non-zero scalar multiple of hh. Thus f∈k[x,y]mPf∈k[x,y]mP.
Once you know that k[x,y]=O(X)k[x,y]=O(X) it is clear that i∗i∗ is an isomorphism. But now their is an equivalence of categories between affine varieties over kk and finitely generated integral domains over kk, and so if XX is affine the map ii has to be an isomorphism too.