Why is the degree of a rational map of projective curves equal to the degree of the homogeneous polynomials?

Let $C_1 \subseteq \mathbb{P}^m$ and $C_2 \subseteq \mathbb{P}^n$ be projective curves, and let $\phi : C_1 \rightarrow C_2$ be a nonconstant rational map given by $\phi = \left[ f_1, \ldots, f_n \right]$ for homogeneous polynomials $f_i \in K[X_0, X_1,\ldots , X_m]$ having the same degree $\deg(f_i) = d$.

The degree of the rational map $\phi$ is defined as

\begin{align*} \deg(\phi) = [ K(C_1) : \phi^* K(C_2)], \end{align*}

where $\phi^* : K(C_2) \rightarrow K(C_1)$ is the injection of function fields given by precomposition: $\phi^* (f) = f \circ \phi$.

In many cases it seems that $\deg(\phi) = d$, but I've been unable to find a proof of this or even conditions on when this will occur. Can someone give some insight into this? Even with specific examples, I can't seem to find a basis for $K(C_1)$ as a vector space over $\phi^*K(C_2)$ to even compute the degree.

Solution 1:

The fact is that deg$(\phi)$ can be anything between $1$ and $d^n$. $1$ is the birational case, and $d^n$ if the sequence $f_0,\cdots,f_n$ is a regular sequence. For example degree of $(s^5:s^3t^2:s^2t^3:t^5):\mathbb{P}^1\to \mathbb{P}^3$ is $1$ and degree of $(s^6:s^4t^2:s^2t^4:t^6):\mathbb{P}^1\to \mathbb{P}^3$ is $2$. I checked the answers in Macaulay2 by packages RationalMaps and Cremona.