If $n\geq 2$, why is $k[\mathbb{A}^n\setminus\{p\}]=k[\mathbb{A}^n]$?

Question

Assuming $n\geq 2$, why is the coordinate ring of affine $n$-space over an algebraically closed field $k$ unchanged if we delete a point? That is, if $p\in \mathbb{A}^n$, why does $k[\mathbb{A}^n\setminus\{p\}]=k[\mathbb{A}^n]$?

This detail seems skimmed over, but it surprised me since I know that $\mathbb{A}^n$ and $\mathbb{A}^n\setminus\{0\}$ are not homeomorphic when $n\geq 2$, so why does the result hold for the coordinate ring? Is there a reference where this is proven? Thanks.

Answer 1

It suffices to consider $p=0$ (since $\mathrm{Aut}(\mathbb{A}^n)$ acts transitively on $\mathbb{A}^n$). Now use that $k[\mathbb{A}^n \setminus \{0\}] = \bigcap_{i=1}^{n} k[x_1,\dotsc,x_n][x_i^{-1}]$ and calculate directly that this is $k[x_1,\dotsc,x_n]=k[\mathbb{A}^n]$. By the way, in more general situations one may use algebraic Hartogs's Lemma (SE/308668).