Question about intersection of coordinate rings

[MathLearner123]

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 $\mathcal{O}(D(X_i))$?)

 

[martinbn]

$D(f)$ 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.