# Degree $1$ birational map of $\mathbb{P}_k^n$

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 $$1$$?

• So a birational map of degree $1$ from $\mathbb{P}^n\to\mathbb{P}^n$ is given by one formula, right? – 6666 Jan 5 at 3:36
• Also why does an automorphism have to be of degree $1$? – 6666 Jan 5 at 4:15
• 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. – KReiser Jan 6 at 9:37