I've a rather simple question that is strucking me, perhaps because I'm a newbie in algebraic geometry.
If we have a projective irreducible non-singular curve $C\subseteq \mathbf{P}^n_k$ where $k$ is an algebraically closed field, and a rational map $f:C\dashrightarrow k$, I'd like to prove that this function defines a regular map $\phi : C\longrightarrow \mathbf{P}^1_k$.
Actually I think I got it: there is an open set $U_f$ for every point $x$ of which we can write $f(x)=F(x)/G(x)$ for some homogeneous polynomials $F,G$ of the same degree. So we should put $\phi (x)=[G(x) : F(x)]$ if $x\in U_f$ and $\phi(x)=[0 : 1]$ if $x\notin U_f$.
Now, the questions are:
How can I prove that $\phi$ is really regular over $C$? It's obviously regular in $U_f$ put what if $p\notin U_f$? I thought about this: since $p\in \mathbf{P}^n_k$ there are homogeneous coordinates $p=[z_0 : \cdots : z_n]$ such that $z_j\neq 0$ for some $j$ and $p\in C\setminus U_f$, that's a closed set; so there's an homogeous polynomial $h$ such that $h(p)=0$ (I suspect that $h$ is a multiple of $G$). Finally we can state $\phi(p)=[0 : 1] = [h(p) : X_j]$, being $X_j (p)= z_j\neq 0$. This proves that $\phi$ is polynomial over the all $C$. Is this correct or I'm missing something?
There a thing that I can't understand about this way of proceeding. If we take $C=V(X_1^2+X_2^2-X_0^2)\subseteq \mathbf{P}^2_k$ and the function $f(X_0,X_1,X_2)=(X_2-X_0)/X_1$, all that we can say is that $$\phi(x_0,x_1,x_2)=\begin{cases} [x_1 : x_2 -x_0] &\mbox{if } x_1\neq 0, x_2\neq \pm x_0 \\ [x_2+x_0 : - x_1] & \mbox{if } x_2\neq -x_0, x_1\neq 0 \\ [0 : 1] &\mbox{otherwise}\end{cases}$$ Why I don't have the situation as before? In fact $f$ is regular at $p=[ 0 : 1 : 1]$ because we have $$\frac{X_2 - X_0}{X_1}=\frac{X_2-X_0}{X_1}\frac{X_2+X_2}{X_2+X_0}=-\frac{X_1}{X_2+X_0}$$ where is defined, BUT $f$ can be evaluated in $p$ only by this latter form. So there're are not polynomials $F,G$ such that $f=F/G$ everywhere $f$ is regular as I thougth before! What I'm missing?
I wonder if all of this could work for a projective variety $X$ of arbitrary dimension... Where do I use that $C$ has dimension $1$ in the preceeding argument?
Thank you very much if you can help!
EDIT
Even if the argument given in 1) doesn't seems completely right to me, thanks to the comments I noticed that 1) holds for the example I made in 2): just define $\phi(x_0,x_1,x_2)=[x_0+x_2 : -x_1]$ over $\Gamma\setminus\{[1:0:-1]\}$ and $[0:1]$ in $ [1:0:-1]$. So the open set I was claiming to not exists in reality there is: $\Gamma\setminus\{[1:0:-1]\}$! I think it's only question of local representation.