Classification of automorphisms of projective space

Question

Let $k$ be a field, n a positive integer.

Vakil's notes, 17.4.B: Show that all the automorphisms of the projective scheme $P_k^n$ correspond to $(n+1)\times(n+1)$ invertible matrices over k, modulo scalars.

His hint is to show that $f^\star \mathcal{O}(1) \cong \mathcal{O}(1).$ (f is the automorphism. I don't if $\mathcal{O}(1)$ is the conventional notation; if it's unclear, it's an invertible sheaf over $P_k^n$) I can show what he wants assuming this, but can someone help me find a clean way to show this?

Answer 1

An automorphism of $\mathbb{P}^n_k$ induces an automorphism of the Picard group $\text{Pic}(\mathbb{P}^n_k) \cong \mathbb{Z}$. Such an automorphism must send the generator $\mathcal{O}(1)$ to a generator. Since the only two generators of $\mathbb{Z}$ are $1$ and $-1$, $f^*(\mathcal{O}(1))$ must be $\mathcal{O}(1)$ or $\mathcal{O}(-1)$. But $\mathcal{O}(-1)$ has no nonzero global sections, so it cannot be the pullback of $\mathcal{O}(1)$ (recall that $\mathcal{O}(1)$ pulls back to a line bundle together with $n+1$ global sections which have no common zero).

Answer 2

Well, $f^*(\mathcal{O}(1))$ must be a line bundle on $\mathbb{P}^n$. In fact, $f^*$ gives a group automorphism of $\text{Pic}(\mathbb{P}^n) \cong \mathbb{Z}$, with inverse $(f^{-1})^*$. Thus, $f^*(\mathcal{O}(1))$ must be a generator of $\text{Pic}(\mathbb{P}^n)$, either $\mathcal{O}(1)$ or $\mathcal{O}(-1)$. But $f^*$ is also an automorphism on the space of global sections, again with inverse $(f^{-1})^*$. Since $\mathcal{O}(1)$ has an $(n+1)$-dimensional vector space of global sections, but $\mathcal{O}(-1)$ has no non-zero global sections, it is impossible for $f^*(\mathcal{O}(1))$ to be $\mathcal{O}(-1)$.