Degree of $f:\mathbb{P}^1_k\rightarrow \mathbb{P}^1_k$ Let $k$ be an algebrically closed field and consider $\mathbb{P}^1_k$ the $1$-dimensional projective space over $k$.
My question is the following:

Let consider $f:\mathbb{P}^1_k\rightarrow \mathbb{P}^1_k$,
  what can be the possible degree of this map?

I answered to myself that if $f:X \rightarrow Y$ is a morphism of curves, $\deg f =[K(X):K(Y)] $ so in our case $\deg f $ can be just equal to 1 and it has to be an isomorphism but it seems to me somehow controintuitive (I think to $S^1 \simeq \mathbb{P}^1_\mathbb{R} $ that admitt TOPOLOGICALLY coverings of arbitrary degree by itself).
Consequence: The same proceedings ($[K(X):K(X)]=1$) can be generalize to generic maps $f:X \rightarrow X$, where $X$ is a curve, saying that this map can only be an isomorphism... Is this true?
 A: Given a dominant morphism $f:X\to Y$ of varieties  (or more generally of integral locally noetherian schemes) the degree of $f$ is the degree of the corresponding field extension $f^*:K(Y)\to K(X)$.
For example the degree of $$f_n:\mathbb P^1_k\to \mathbb P^1_k: [x:y]\mapsto [x^n:y^n]$$ is the degree of the extension of fields $$f^*_n:k(t)\to k(t):t\mapsto t^n$$ where  $t=\frac yx$.
That last degree is $n$ and so we can obtain morphisms of any given degree: $deg (f_n)=n$.  
Optional complement
Notice that if $k$ is a field of characteristic $p\gt 0$ the Frobenius morphism $$Frob=f_p:\mathbb P^1_k\to \mathbb P^1_k: [x:y]\mapsto [x^p:y^p]$$  has degree $p$ (as noted above) but is  bijective, and even a homeomorphism.
This proves that in general situations we can't define degrees by naïvely counting points in fibers: all  fibers of the Frobenius morphisms consist in just one point and yet the morphism has degree $p$.
Notice how strange it is that a morphism which is a homeomorphism may nevertheless not be a birational map. 
