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I'm trying to solve exercise 1.3.8 from Hartshorne's Algebraic Geometry:

Let $ H_{i} $ and $ H_{j} $ be the hyperplanes in $ \mathbb{P}^n $ defined by $ x_i = 0 $ and $ x_j = 0 $ with $ i \neq j $. I Show that any regular function on $\Bbb{P}^n - (H_i \cap H_j)$ is constant.

I found a solution set online claiming that $ \mathcal{O}(\mathbb{P}^n - H_i) \cong (k[x_0,\ldots,x_n]_{x_i})_{0} $. This is used to show that any regular function is equal to $ f_i/x_i^a = f_j/x_j^b $ and then this forces the function to be constant.

I don't know where the first isomorphism comes from. I can't find such a theorem in the book up to the end of section 3 in chapter 1. The difficulty I'm having is that $ \mathbb{P}^n - H_i $ is not closed, so it is not a projective variety. The theorems up until that point in the book are only applicable to projective varieties as far as I can tell.

I feel that I'm missing something obvious here. Any help would be appreciated.

PeterM
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1 Answers1

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This isomorphism appears in the first paragraph of the proof of Theorem 3.4.

More generally, if $Y$ is projective, $Y - \{f=0\}$ is isomorphic to the affine variety $\operatorname{Spec} (({S(Y)}_f)_0)$. We have $K(Y)=S(Y)_0$, and a function is defined on all of $Y-\{f = 0\}$ exactly when its denominator is a power of $f$.

Andrew Dudzik
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  • How does the isomorphism $\mathcal{O}(\mathbf{P}^{n}-H_{i})\cong(k[x_{0},...,x_{n}]{x{i}}){0}$ show up in the first paragraph of Theorem 3.4? In that paragraph, I see the isomorphism $k[y{1},...,y_{n}]\cong k[x_{0},...,x_{n}]{(x{i})}$. I see that $\mathcal{O}(\mathbf{P}^{n}-H_{i})\cong k[y_{1},...,y_{n}]$, but I don't see how $(k[x_{0},...,x_{n}]{x{i}}){0}\cong k[x{0},...,x_{n}]{(x{i})}$. Could you clarify? – JDZ Aug 27 '19 at 03:37
  • @JDZ Your last equation is true by definition: the homogeneous localization $S_{(x_i)}$ is defined to be $(S_{x_i})_0$, the set of degree zero elements of the regular localization. (this post is over 5 years old, so let me know if I've missed the point here) – Andrew Dudzik Aug 27 '19 at 10:01
  • $S_{(x_{i})}$ is by definition the localization of $S$ at the complement of the prime ideal $(x_{i})$ in $S$ (cf. Hartshorne p. xvi of the introduction), whereas $S_{((x_{i}))}$ is by definition $(S_{x_i}){0}$ (cf. Hartshorne p. 18 in the paragraph above theorem 3.4). I think that Hartshorne mistakenly wrote $S{(x_{i})}$ in the first paragraph of the proof of theorem 3.4, when he really meant $S_{((x_{i}))}$. It is a slight notational difference between two different things, which is unfortunate. – JDZ Aug 27 '19 at 21:01
  • @JDZ Check the last sentence in the paragraph before Theorem 3.4. – user5826 Sep 12 '20 at 17:30