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I have trouble understanding what the coordinate ring of an open affine variety, i.e. the ring of regular functions on that variety, is (in the classical sense). Let $k$ be algebraically closed.

If $X \subseteq \mathbb A^n$ is a closed affine variety, then the coordinate ring is defined as $k[X] := k[x_1, \dots, x_n] / I(X)$ where $I(X) = (f_1, \dots, f_m)$ is the finitely generated ideal of polynomials vanishing on $X$.

If $D(f) \subseteq \mathbb A^n$ is a distinguished open set for some $f \in k[x_1, \dots, x_n]$, then $k[D(f)] = k[x_1, \dots, x_n, f^{-1}]$ is the localisation at $f$.

Now for a general open affine variety $U \subseteq \mathbb A^n$, we can write them as $U = \mathbb A^n \setminus X$ for a closed variety $X = V(f_1, \dots, f_m)$ where $X$ is the vanishing set of $f_1, \dots, f_m \in k[x_1, \dots, x_n]$. How can I construct $k[U]$? Is it simply the $k$-algebra generated by the $k[D(f_i)]$ for $i = 1, \dots, m$, i.e. $k[U] = k[x_1, \dots, x_n, f_1^{-1}, \dots, f_m^{-1}]$?

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  • $\begingroup$ It is also called "ring of regular functions", as regular functions on $Y$ are polynomials -we're using only algebraic operations in algebraic geometry- in coordinates on $Y$. Taking a closed subvariety $Y\subseteq X$ corresponds to quotienting by functions on $X$ vanishing on $Y$, as such functions give no information/do not depend on coordinates on $Y$. All that is left are regular functions of coordinates, all of them, as we assume we had all such function on $X$, and those on $Y$ are just restrictions of those on $X$ -and by quotienting we prune out the indeterminacy. Last question: Yes. $\endgroup$
    – plm
    Commented Jun 12 at 12:51
  • $\begingroup$ [2/2] I should add that functions on the complement of $Y$ contain all functions defined on $X$, as a regular function is determined by its values on an open set, at least with an algebraically closed base field; but we can define more regular functions on $X\setminus Y$ than on $X$: all the functions vanishing only on $Y$ now have regular inverse, so we add the inverse of $f_i$ for each $i$, and those are all the extra regular functions on $U$ compared to $X$, so we get your formula for $k[U]$. $\endgroup$
    – plm
    Commented Jun 12 at 13:05

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If you have an open variety $U \subsetneq {\mathbb A}^n$ which is an affine variety, then the complement $H:= {\mathbb A}^n \setminus U$ is a hypersurface and thus the zero set of a polynomial $f$. It follows that the coordinate ring of $U$ is $k[x_1,\ldots,x_n,f^{-1}]$.

The claim about the complement is a consequence of the following result.

If $X$ is a normal variety and $U \subset X$ an open subset such that the complement $X \setminus U$ has codimension $\geq 2$ in $X$, then every regular function on $U$ extends to a regular function on $X$.

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