# 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?

• Your purported formula for $\deg f$ looks Highly Suspicious, since $K(Y)$ isn't generally a subfield of $K(X)$; particularly, the right-hand side doesn't depend on $f$. Probably you want something like $[K(X) : f^{*}K(Y)]$, the degree of $K(X)$ over the subfield of functions on $X$ that factor through $f$. (Over the real or complex numbers, the $n$th power map $[z, 1] \mapsto [z^{n}, 1]$ has degree $n$, as your intuition suggests.) Nov 6, 2015 at 11:05
• Sorry, I'm probabily wrong, but is not a consequence of the fatct that $f$ is dominant? Nov 6, 2015 at 11:50
• @Andrew: Joseph's degree formula is perfectly correct ALWAYS . As soon as $f:X\to Y$ is dominant, which for curves means nonconstant, $f$ induces a field extension $K(X)\to K(Y)$ whose degree is the degree of $f$. This, and not counting points in fibers, is the only correct general definition of degree in algebraic geometry . By the way, a field extension is a morphism of fields, necessarily injective, but not an inclusion in general. Nov 6, 2015 at 12:41
• Dear Joseph: you are not wrong, but absolutely right ! Nov 6, 2015 at 12:42
• Dear @Andrew, yes the degree of $f:X \to X$ is indeed the degree $[K(X):f^*K(X)]$. The subtle point is that the inclusion $f^*(K(X))\subset K(X)$ will be strict in general. In other words the field $K(X)$ has strict subfields isomorphic to itself : $f^*(K(X))\subsetneq K(X)$ BUT $f^*(K(X))\cong K(X)$ as abstract fields. In my example below we have $f_n^*(K(\mathbb P^1 ))=k(t^n)\subsetneq K(\mathbb P^1 )=k(t)$ BUT $k(t^n)\cong k(t)$ as abstract fields . This is analogous to the statement $2\mathbb Z\subsetneq\mathbb Z$ BUT the groups $2\mathbb Z$ and $\mathbb Z$ are isomorphic as groups. Nov 6, 2015 at 23:56

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: $deg (f_n)=n$.
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 $p$.
• The strangeness in the last line of this answer is fortunately absent in the case of an algebraically closed field of characteristic zero: a bijective morphism $X\to Y$ between two varieties $X,Y$ with $Y$ normal will then be an isomorphism. This is astonishingly difficult to prove and involves Zariski's Main theorem. See this thread. Nov 7, 2015 at 0:07