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Consider $\mathbb{P}^n$ as an algebraic variety over $\mathbb{C}$. I'm trying to solve the following exercise:

Prove that $\text{Aut}(\mathbb{P}^n)=\text{PGL}(n+1,\mathbb{C})$

[hint: prove that automorphisms of $\mathbb{P}^n$ send hyperplanes into hyperplanes]

Let $H:=Z(L)$ be a hyperplane, where $L$ is a linear form. I've noticed that if $f\in \text{Aut}(\mathbb{P}^n)$, then $f^{-1}(H)=Z(L\circ f)$. Taking a representation $f=(F_0:...:F_n)$ with $F_i$'s homogeneous of same degree $d$, then $L\circ f$ gives us a polynomial of degree $d$.

If $H=Z(x_i)$, and assuming the claim from the hint, then $Z(x_i\circ f)=Z(F_i)$ is a hyperplane, i.e., $d=\deg(F_i)=1$. Since the $F_i$'s can be rescaled by a non-zero constant, then $f\in\text{PGL}(n+1,\mathbb{C})$.

Now I can't see any intuitive reason why the claim should be true. I'm sure it must be some very particular feature of $\mathbb{P}^n$, but I don't know what it is.

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rmdmc89
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    Sorry, marked this as a duplicate of the wrong question the first time around. This is a duplicate of https://math.stackexchange.com/questions/150605/classification-of-automorphisms-of-projective-space/150614 and should be marked as such. – KReiser Jan 24 '20 at 01:16
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    @KReiser, I'm not familiar with schemes, bundles etc, so the post you mentioned doesn't help me, unfortunately – rmdmc89 Jan 24 '20 at 01:33
  • You don't need any knowledge of schemes for this one - just what an invertible sheaf is. Do you know that? – KReiser Jan 24 '20 at 05:45
  • Do you have access to Bézout’s Theorem? – Jesko Hüttenhain Jan 25 '20 at 22:31

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