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It is well-known that $X=\mathbb C^2 \setminus \{ (0,0) \}$ is not an affine variety. Perhaps the simplest proof of this fact consists of showing that every regular function on $X$ extends to a regular function on $\mathbb C^2$. This is easy to do with elementary algebraic tools only. This approach is sometimes described as an algebraic version of Hartogs's principle. However, we don't actually have to invoke complex analysis. I am wondering if it is really possible to do it this way, though. Certainly Hartogs's theorem would tell me that every regular function on $X$ extends to a holomorphic function on $\mathbb C^2$. Is there some general principle that guarantees that this extension is a regular function?

Blazej
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    I'm not sure if you're asking for an elementary argument to avoid Hartogs (we can just write down the Cauchy integral formula for this special case) or whether a holomorphic function that is regular away from the origin is everywhere regular. – Ted Shifrin Nov 18 '19 at 02:17
  • I am asking the latter. I do know that in this simple geometric setup extension may be constructed by contour integrals and I would like to know if it is possible to argue that these formulas produce regular functions. – Blazej Nov 18 '19 at 02:51
  • Here is a purely algebraic explanation avoiding complex analysis and thus not invoking Hartogs's theorem. It is based on $\mathbb C^2$ being a normal algebraic variety, which in your case just means that the ring $\mathbb C[X,Y]$ of regular functions on $\mathbb C^2$ is integrally closed. – Georges Elencwajg Nov 18 '19 at 13:11
  • See also there. – Georges Elencwajg Nov 18 '19 at 13:39
  • I am aware of these proofs. In fact proof using complex analysis is precisely what I am after. – Blazej Nov 19 '19 at 03:11
  • So we have a holomorphic function on $\Bbb C^2$ (or, indeed, on any open subset) that is a polynomial on $\Bbb C^2-{0}$ . It follows from the identity principle that it agrees with the polynomial at the origin. – Ted Shifrin Nov 20 '19 at 18:08
  • What if I instisted on not using the fact that any regular function on $\mathbb C^2 \setminus { 0 }$ is a polynomial? – Blazej Nov 21 '19 at 00:51

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