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I have some major confusion about about the process of generalizing affine varieties to quasi projective varieties. We work over an algebraically closed field $\mathbb{k}$. A quasi projective variety is an open subset of a closed set in projective space. An affine variety $X \subset \mathbb{A}^n$ is a quasi projective variety since we have a homeomorphism from $\mathbb{A}^n$ to an affine piece $\mathbb{A}^n_i$ of $n$ dimensional projective space. Now I am not sure exactly how we transfer the additional structure that a variety has besides its topological structure.

The ring of regular functions $\mathbb{k}[X]$ was defined for an affine variety to be functions $f: X\to \mathbb{k}$ that are given by polynomials, or, for an irreducible affine variety, as elements of the field of fractions of $\mathbb{k}[X]$ such that for each point there is a representation such that the denominator is nonzero at that point. It was proven that these two definitions are equivalent. We define regular functions on quasi projective varieties as by saying a function $f:X \to \mathbb{k}$ is regular if there are homogeneous polynomials $p$ and $q$ in $n+1$ variables of the same degree such that at every $x\in X$ we have $f(x) = p(x)/(q(x)$. Now to show that this is the same definition as for an affine variety I don't even understand what one would try to do. There is supposedly a proof in Shafarevich which makes no sense to me.

One of the confusing parts for me is that we could "embed" an affine variety into projective space in many ways. Is there some formal way of saying that the "particular embedding doesn't matter"? To do this I assume we would want to show that any two "embeddings" are isomorphic. I'm a little lost here. What exactly should we define "embedding" to mean? And how do we show that the algebraic variety structure on the embedding is "the same" as the structure on the origional affine variety?

For example, for one of the homework problems I am supposed to compute $\mathbb{k}[X]$ for the quasi projective variety $\mathbb{A}^2 - 0$. I don't think this even makes sense unless we view $\mathbb{A}^2 - 0$ as a subset of some projective space. I don't understand if I can justificably choose "any" embedding or not.

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  • $\begingroup$ One way of making concrete the additional structure that a variety has is to attach the sheaf of regular functions to it and get a locally ringed space. $\endgroup$ – Zhen Lin Feb 23 '14 at 0:38
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Let $X$ be an affine variety and let $X'$ be its embedding to projective space (by embedding affine space to $x_0 = 0$). Then given a regular function on $X$ (say $f$) you get a regular function on $X'$ by homogenizing $f$ and then dividing by $x_0$ to the degree of $f$. Conversely given a regular function $g\over h$ on $X'$ you get a regular function on $X$ by setting $x_0=1$. This correspondence gives the isomorphism between ring of regular functions of $X$ and that of $X'$. Since we embed isomorphically, it doesn't matter how we embed (you can find an isomorphism between two embedding by composing the inverse of one embedding with the other embedding).

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  • $\begingroup$ Thank you, I guess the homeomorphism between $\mathbb{A}^n$ and $\mathbb{A}^n_i$ is an isomorphism of varieties almost from the definition of regular map. This whole setup is still confusing to me but its starting to make more sense. Can you comment on why given a regular map $\frac{g}{h}$ on $X'$ setting $x_0 = 1$ gives a regular map on $X$? $\endgroup$ – Seth Feb 23 '14 at 21:03
  • $\begingroup$ @Seth: The correspondence between $X$ and $X'$ is given by: $(a_1,a_2,...,a_n)\in X$ iff $(1,a_1,a_2,...,a_n)\in X'$. $\endgroup$ – user129924 Mar 4 '14 at 10:46

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