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Let $X=\mathbb{P}^2$ and $C=V(x^2+y^2-z^2)$ an irreducible conic. Then the Riemann-Roch space associated to $C$ by definition is $\mathcal{L}(C)=\{ f \in k(X)^{*} \vert div f + C \geq 0 \} \cup \{0 \}=\{\frac{F}{x^2+y^2-z^2} \vert F \text{ is homogeneous of degree } 2 \} \cup \{0 \}$. That is, $\mathcal{L}(C)$ is a vector space with basis $\{ \frac{x^2}{x^2+y^2-z^2}, \frac{xy}{x^2+y^2-z^2}, \frac{y^2}{x^2+y^2-z^2}, \frac{yz}{x^2+y^2-z^2}, \frac{xz}{x^2+y^2-z^2}, \frac{z^2}{x^2+y^2-z^2} \}$. For short, we can name the basis elements $\varphi_0, ..., \varphi_5$. There is a rational map associated to $\mathcal{L}(C)$. That is, $ \varphi_{|C|}: X \to \mathbb{P}^5$, defined as $\varphi_{|C|}(x)=[\varphi_0(x): ... : \varphi_5(x)]$. At first glance, it seems that $\varphi_{|C|}$ is not defined exactly at points $x \in C$. But then the linear system $|C|$ is base point free and there's a theorem which says that is equivalent to $\varphi_{|C|}$ being regular, so defined everywhere.

What am I badly missing here?

Kristina
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    Note that the rational map you have written is equal to the one you would get if you cleared all the (identical) denominators in the $\varphi_i$. The latter map extends $\varphi_{|C|}$ to a regular map $\mathbf P^2 \rightarrow \mathbf P^5$. (This regular map is famous: do you know its name?) –  Apr 06 '15 at 21:58
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    By the way, this question and its close relatives are extremely common! See e.g. http://math.stackexchange.com/questions/1215990/any-rational-map-can-be-extended-to-codimension-one/ for a very similar question from just a few days ago. So rest assured that you are having the same learning experience as many others. –  Apr 06 '15 at 22:18
  • Dear @AsalBeagDubh, I know I should get the Veronese embedding. But that's the thing - I don't know why I can clear those denominators... Also, I've noticed that question from few days ago, but I'll read it more carefully. Thanks for letting me know. – Kristina Apr 06 '15 at 22:25
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    You can clear the denominators because of the way homogenous coordinates work: for each point $p \in \mathbf P^2 \setminus C$, the value of $x^2+y^2-z^2$ at $p$ is some fixed nonzero number, $a$ say. So the image of $p$ in $\mathbf P^5$ is the same whether we calculate it via the Veronese or via your original formula: the difference is just a bunch of $a$'s appearing in the denominators of each entry, which doesn't change the point in projective space. –  Apr 06 '15 at 22:32
  • Of course. But what about $p \in C$? Or do you mean, first clear the denominators and get a regular map $\mathbb{P}^2 \setminus C \to \mathbb{P}^5$ and then just extend to a Veronese embedding $\mathbb{P}^2 \to \mathbb{P}^5$? – Kristina Apr 06 '15 at 22:49
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    Yes, I mean the latter. Clearing the denominators shows that on $\mathbf P^2 \setminus C$ your map is just the restriction of the Veronese, so then you know how to extend it. –  Apr 06 '15 at 22:54
  • Well, thank you then! – Kristina Apr 06 '15 at 22:58

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