2
$\begingroup$

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:

  1. 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?

  2. 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?

  3. 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.

$\endgroup$
  • $\begingroup$ Sorry, I added some important specifics to the curve. $\endgroup$ – Caligula Jan 26 '15 at 23:33
  • $\begingroup$ This stuff is definitely worked out in detail in one of the first two chapters of Shafarevich. I don't have time to go through the details right now, but here are some quick comments: for 2), yes, this is fine. You will always need to use different representations of the rational function at different points on the curve, since the set of poles of any given nonconstant rational function is a hypersurface in $P^n$, hence will intersect the curve. 3. No, this definitely doesn't work in higher dimensions! It really is special to curves. Try the rational function $Y/X$ on $P^2$ --- $\endgroup$ – user64687 Jan 27 '15 at 8:44
  • $\begingroup$ how should it be extended to the point $[0,0,1]$? $\endgroup$ – user64687 Jan 27 '15 at 8:45
  • $\begingroup$ Thank you @AsalBeagDubh, also for references. I think I'm missing something rather trivial about definition: maybe because I'm mistaking global and local. It's true that if $f\in k(C)$ is a rational function, then $f$ equals the quotient of two polynomial on a WHOLE open set? I think this must be wrong in some sense, because for the $f$ as above in 2) this open set in not fixed but it changes with the point. $\endgroup$ – Caligula Jan 27 '15 at 9:33
  • $\begingroup$ Regarding $Y/X$ over $\mathbf{P}^2$, I'm not seeing it, sorry. I obviously feel there is a problem to extend $f$ when $Y=X=0$ but how do this relate to the fact $\mathbf{P}^2$ has got dimension $>1$? $\endgroup$ – Caligula Jan 27 '15 at 9:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.