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I wonder why the degree $1$ birational maps of $\mathbb{P}_k^n$ are the automorphisms of $\mathbb{P}_k^n$? In particular, why are they defined everywhere on $\mathbb{P}_k^n$?

I know each degree $1$ birational maps of $\mathbb{P}_k^n$ is of the form $\phi:=[f_0:...:f_n]$, where $f_i$ is a homogeneous linear polynomial. I know $\phi$ is defined everywhere iff the associated matrix (coefficients of $f_i$) of $\phi$ is corresponding to a matrix in $\text{PGL}_{n+1}(k)$, i.e. $\text{Aut}(\mathbb{P}_k^n)=\text{PGL}_{n+1}(k)$. But how to see the set of degree $1$ birational maps of $\mathbb{P}_k^n$ is $\text{Aut}(\mathbb{P}_k^n)$?

Also why does an automorphism of $\mathbb{P}^n$ have to be of degree $1$?

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The key to this is the matrix of linear forms. If the matrix is not full rank, then the image of the map associated to this matrix of linear forms is supported inside some closed linear subvariety, which implies it is not birational. So birationality is equivalent to the matrix being full rank, which is equivalent to it being defined everywhere: the points where the map associated to the matrix isn't defined is exactly the projectivization of the kernel of this matrix.

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  • $\begingroup$ Can you explain why the closed linear subvariety implies it’s not birational? Why can’t a variety be birational to its closed linear sub variety? $\endgroup$ – 6666 Jan 3 at 16:43
  • $\begingroup$ Birationality preserves dimension. Closed (proper) linear subvarieties of projective space have lower dimension. $\endgroup$ – KReiser Jan 3 at 20:26
  • $\begingroup$ So a birational map of degree $1$ from $\mathbb{P}^n\to\mathbb{P}^n$ is given by one formula, right? $\endgroup$ – 6666 Jan 5 at 3:36
  • $\begingroup$ Also why does an automorphism have to be of degree $1$? $\endgroup$ – 6666 Jan 5 at 4:15
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    $\begingroup$ You still need to be careful - you want to say that the maximal domain of definition of this rational morphism is $\Bbb P^n$, and on this domain, it's represented by an element of $PGL_{n+1}(k)$. If you do this, you can say the formula is unique up to scaling - if you allow yourself to work on a smaller open subset, this may not be the case. Remember, rational maps are equivalence classes, so you need to deal with that aspect of things. $\endgroup$ – KReiser Jan 6 at 9:37

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