Fix two integers $n \geq 1$ and $d \geq 2$. Let $X_{n,d} \subset \mathbb{P}^{n+1}_k$ be the Fermat hypersurface defined by the equation $$ F_{n,d}: X_0^{d}+ \cdots + X_{n+1}^{d} = 0. $$ One can check that it is nonsingular of dimension $n$. In the exam (EDIT: This is from an exam paper in 2017, so this post is not cheating), I was asked to consider the following problem:
Question: Let $p \in X_{n,d}$ be any point. Show that $X_{n,d} \smallsetminus \{p\}$ is not projective. When is it affine?
I managed to see why it is not projective, but got stuck on considering its affineness.
My attempts: When $n \geq 2$, the singleton $\{p\}$ has codimension $\geq 2$ in $X_{n,d}$. Hence by the Hartog extension theorem, the inclusion $i: X_{n,d} \smallsetminus \{p\} \hookrightarrow X_{n,d}$ induces an isomorphism $$ i^{\sharp}: \Gamma(X_{n,d}, \mathcal{O}_{X_{n,d}}) \rightarrow \Gamma(X_{n,d} \smallsetminus \{p\}, \mathcal{O}_{X_{n,d} \smallsetminus \{p\}}) $$ so the global section of $X_{n,d} \smallsetminus \{p\}$ is $k$. If it were affine, then its dimension could be calculated by the Krull dimension of the global section, hence its dimension would be zero, which is absurd. Hence it is not affine.
However, when $n=1$, the Hartog extension theorem is not applicable. Then I do not know what to do. When $n=1$ and $d=2$, if we use the "huge input" that every nonsingular conic in $\mathbb{P}^2$ is isomorphic to $\mathbb{P}^1$, then we see $X_{1,2} \smallsetminus \{p\}$ is affine, isomorphic to $\mathbb{A}^1$. But how about the general degree $d > 2$?
(And as for the "huge input", I do not want to use it since the only proof of it I know is to see both the nonsingular conic in $\mathbb{P}^2$ and $\mathbb{P}^1$ has genus zero, and all genus zero curves are isomorphic by Riemann-Roch. So can we get rid of this input or can we use a more elementary approach for this "huge input"?)
Thank you so much for answering and commenting! Happy new year!
