A Question about intersection of coordinate rings

MathLearner123
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I want to prove that ##\mathcal{O}_{\mathbb{A}^n}(\mathbb{A}^n \setminus \{0\}) = k[X_1, \ldots, X_n]## for ##n \ge 2## and some user on another forum gave this proof:

"We know that ##\mathcal{O}_{\mathbb{A}^n}(\mathbb{A}^n \setminus \{0\}) = \bigcap_{i = 1}^n \mathcal{O}(D(X_i))##, where ##D(f) = \{x \in \mathbb{A}^n : f(x) \ne 0\}##. Now ##\mathcal{O}(D(X_i)) = k[X_1, \ldots,X_n,\frac{1}{X_i}]## and ##\mathcal{O}(D(X_i)) \bigcap \mathcal{O}(D(X_j)) = k[X_1, \ldots, X_n]## for every ##i \ne j##.

This implies that ##\bigcap_{i = 1}^n \mathcal{O}(D(X_i)) = k[X_1,\ldots,X_n]##, so ##\mathcal{O}_{\mathbb{A}^n}(\mathbb{A}^n \setminus \{0\}) = k[X_1,\ldots, X_n]##."

I don't understand some notations. I know that ##\mathbb{A}^n \setminus\{0\} = \bigcup_{i=1}^n D(X_i)## and that ##\mathcal{O}_{\mathbb{A}^n}(\mathbb{A}^n \setminus \{0\}) = \{f : \mathbb{A}^n \setminus \{0\} \to k : f \text{ is regular on U}\}## and for every ##i \in \{1, 2, \ldots, n\}## we have ##\mathcal{O}(D(X_i)) = k[X_1,\ldots,X_n]/(I(X_i))## where ##I(X_i) = \{f \in k[X_1,\ldots,X_n] : f(x) = 0 \text{ for any } x \in (X_i)\}##. Now I don't understand why we can intersect those quotient rings. I know that we can interpret ##\mathcal{O}(D(X_i))## as restrictions of polynomials on ##D(X_i)## but I still don't understand how we can intersect those terms. (And, shouldn't be ##\mathcal{O}_{\mathbb{A}^n}(D(X_i))## instead of ##\mathcal{O}(D(X_i))##?)
 
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##D(f)## is the set on which ##f## is not zero, and ##\mathcal O(D(f))## is the ring of regular functions on that set. It is not a quotient, it is a localization. You can have any powers of ##f## in the denominator. You are confusing them with the zero of an ideal and the regular functions on it. This is why ##\mathcal O(D(X_i))## is ##k[X_1,X_2,\dots,X_n,\frac1{X_i}]## and not the quotient you think. All of these are subrings of the field of rational functions and the intersection is inside it.
 
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