Degree of a divisor for algebraically closed fields vs not closed ones Suppose we have an algebraically closed field
$F$ and we consider the projective space $\mathbb{P}^1$
over $F$. If we consider some divisor 
$D = n_P P + n_Q Q +n_s S$, then we say the degree of $D$ 
is $n_P  + n_Q  +n_s $, the sum of coefficients.
However, if $F$ is not algebraically closed, for example 
a field of finite order, then for a divisor 
$D = n_P P + n_Q Q +n_s S$, where $P,Q,S$ are closed points,
the degree of $D$ is
$$
n_P (\deg P) + n_Q (\deg Q) +n_s (\deg S). 
$$
I was wondering if someone could possibly give me some kind of explanation
for why this is a natural thing to do for fields that are not algebraically closed.
Thank you very much!
 A: An  intuitive reason is that given a point $P\in \mathbb P^1_F$ we want consider it big  in proportion to the bigness of  its residue field $\kappa(P)=\mathcal O_{P}/\mathfrak  m_{P}$ and the measure of that bigness is the dimension $deg (P)=dim_F(\kappa(P))$.
So the degree of the point  $P\in\mathbb A^1_F\subset \mathbb P^1_F$ corresponding to an irreducible polynomial $p(x)\in F[x]$ is the degree of $p(x)$: doesnt that look reasonable?  
But the main technical  reason for this convention is that we want the degree of the divisor $div(\phi)$ of a rational function $\phi(x)\in F(x)$ to be zero.
And this forces the definition of the degree of a point, and hence of a divisor, of  $\mathbb P^1_F$.    
For example take the rational function $\phi(x)=\frac{x^2+1}{x}\in \mathbb R(x)$ .
The corresponding divisor on $\mathbb P^1_\mathbb R$ is $div(\phi)=1.P-1.O-1.\infty$, where $P$ is the point corresponding to the irreducible polynomial $x^2+1\in \mathbb R[x]$ ,  $O$ is the rational point corresponding to the irreducible polynomial $x\in \mathbb R[x]$ and $\infty$ is the rational point at infinity of $\mathbb P^1_\mathbb R$.
It is now perfectly clear that if you want to force $deg( div(\phi))=deg(P)-1-1$ to be zero you must define $deg(P)=deg(x^2+1)=2$ .
