Degree of the negative part of a divisor

Question

Let $K$ be an algebraically closed field (or $\overline{\mathbb{C}(z)}$ for a more precise condition). And let $P \in K[x,y]$ be an irreducible polynomial of degree $m$ with respect to $x$ and degree $n$ with respect to $y$. Consider the function field $R:=Frac(K[x,y]/(P))$ associated to $P$.

I would like to know why $\deg \delta^{-} (x) = n$ and $\deg \delta^{-} (y) = m$. Here, for $t \in R$, $\delta (t)$ denotes the sum $\sum (ord_p(t)).p$ with p running over all places of $R$; and $\delta^- (t)$ is the negative part only of $\delta (t)$; and $\deg \delta(t):= \sum ord_p(t)$. Any help would be greatly appreciated!

Answer 1

First we know that for arbitrary $r\in R$, $R/K(r)$ is an algebraic extension of rational function field $K(r)$ and $deg^-(r)=deg^+(r)=[R:K(r)]$. A proof can be found in Chapter 1 of Stichtenoth's book.

Second,$R =K(x,y)=K(x)[y]=K(y)[x]$ i.e R is a finite extension of $K(x)$ by $y$ where y is a root of $G(T)=P(x,T)\in K(x)[T]$. Since $G(T)$ is an irreducible polynomial of degree $m$, then $[K(x)[y]:K(x)]=m$ which is the degree of $x$. Similarly $R$ is a finite extension of $K(y)$ by $x$. So $[K(y)[x]:K(y)]=deg(H(T))$ where $H(T)=P(T,y)$ is the irreducible polynomial of $x$ over $K(y)$.