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In general, how does one determine a local ring. And in particular, how would one do it for $O_{A}(A $ \ $ \{(0)\})$, where A is 1-dim affine space in $\mathbb{C}$?

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In general, when $X$ is an affine variety and you are interested in $\mathcal O_X(X\setminus Z)$ with $Z=Z(f)$ a hypersurface, then $\mathcal O_X(X\setminus Z)=\mathcal O_X(X)_f$ is the localization of $\mathcal O_X(X)$ at $f$. In your case, $Z=\{ 0\} = Z(x)$, so $\mathcal O_A(A\setminus \{0\})=\mathcal O_A(A)_x = k[x]_x = k[x,x^{-1}].$

However, note that this is not a local ring! It is a localization at $x$, but a local ring is a ring with a unique maximal ideal and $k[x,x^{-1}]$ has more than one maximal ideal.

Note that this is not easily done when $Z$ is not a hypersurface. For example, $\mathcal O_{A^2}(A^2\setminus\{0\})$ is the same as $\mathcal O_{A^2}(A^2)$, namely the polynomial ring in two variables.

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  • $\begingroup$ I do not really understand your point with A^2. Why would they be the same? $\endgroup$ – math1234567 Apr 5 '14 at 21:14
  • $\begingroup$ In larger generality, when $X$ is normal and $Z\subseteq X$ is of codimension greater or equal than two, then $\mathcal O(X)=\mathcal O(X\setminus Z)$. In this concrete case, it means that any regular function on $A^2\setminus\{ 0 \}$ can be extended to a polynomial on $A^2$. Indeed, interpret that function as a rational function $f/g$ - there is no way that $g$ only vanishes at a single point, it would have to vanish at a codimension one subvariety. Hence, the function must be a polynomial. $\endgroup$ – Jesko Hüttenhain Apr 5 '14 at 21:21
  • $\begingroup$ And by $A^2\{0}$, you mean $(A^2\{(0,0})$? $\endgroup$ – math1234567 Apr 6 '14 at 17:49
  • $\begingroup$ Yes, that is what I mean. $\endgroup$ – Jesko Hüttenhain Apr 6 '14 at 17:50
  • $\begingroup$ Sorry for all of the questions, but what does the subscript x mean? (with $O_A(A)_x=k[x]_x$) $\endgroup$ – math1234567 Apr 6 '14 at 19:13

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