Degree of a morphism from a curve to $\mathbb P^1_\mathbb C$: explicit description Let $f:X\to \mathbb P^1_{\mathbb C}$ be a non-constant (i.e. surjective) morphism (of $\mathbb C$-varieties/schemes) from a smooth complex projective curve to the projective line. The degree  of the morphism $f$ is defined "abstractly'' as the degree of the extension between the function fields:
$$d=deg(f):=[\mathbb C (X): \mathbb C(\mathbb P^1)]=[\mathbb C (X): \mathbb C(t)]$$

Now lets embed $X$ is some $\mathbb P^N$ and try to "get our hands dirty'' with the morphism $f$. By looking in local coordinates and in a classical pre-Grothendieck setting one can see that $f$ is given by the datum of $2$-uples of homogeneous polynomials in $N$ variables and of the same degree:
$$f\sim\{F_{0k},F_{1k}\}_k$$
satisfying certain (boring) glueing conditions.
I was wondering if the degree $d$ of the morphism $f$ is equal to the degree of the polinomials $F_{ij}$. Is it true? 
I'm guided by the following example:
$$\mathbb P^1\to\mathbb P^1$$
$$[1:t]\mapsto [1:t^n]$$
$$\infty\mapsto\infty$$
gives a morphism of degree $n$.
 A: I don't think this is true. Consider the example where $X$ is cut out by $X^2+Y^2+Z^2=0$ in $\mathbb{P}^2$ and the map $X\rightarrow \mathbb{P}^1$ given by $[X:Y:Z]\rightarrow [X:Y]$. 
If you fix $[X:Y]\in \mathbb{P}^1$, the preimage will consist of two points (unless $[X:Y]=[1:I]$ or $[1:-I]$), so the map is degree 2. 
In general, I think the degree $d$ of $f$ is going to be the degree of $X$ in $\mathbb{P}^N$ times the degree of the polynomials inducing $f$. I also don't think that every map $X\rightarrow \mathbb{P}^1$ is induced by homogenous polynomials in $N+1$ variables, and I think you can construct counterexamples using my first claim. 
A: I think the simplest counterexample is if you take $C$ to be the standard twisted cubic, viz. the image of the closed embedding $f\colon\mathbb{P}^1\to\mathbb{P}^3$ given by $[U:V] \mapsto [U^3:U^2V:UV^2:V^3]$ (if we call $[T:X:Y:Z]$ the coordinates on $\mathbb{P}^3$, it has equations $TY=X^2$, $XZ=Y^2$, $TZ=XY$, but no matter).  This morphism is an isomorphism onto its image as is easy to see (e.g., its inverse $g_1\colon C\to\mathbb{P}^1$ is $[T:X:Y:Z]\mapsto[T:X]=[X:Y]=[Y:Z]$), but the morphism $g_2\colon [T:X:Y:Z]\mapsto[T:Y]=[X:Z]$ and the morphism $g_3\colon [T:X:Y:Z]\mapsto[T:Z]$ are of degrees respectively $2$ and $3$ as can be seen by composing on the right by $f$.  In fact, $g_1,g_2,g_3$ are given by polynomials of the same degree, but the morphisms in question have three different degrees.
