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!
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 $deg(P)=deg(x^2+1)=2$ .