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I am still on my quest to design a simple example of an ample, globally generated line bundle which is not very ample and also understand the global sections of its smallest, very ample tensor power. Let $\Bbbk$ be some algebraically closed field, Consider $$R = \Bbbk[x,y,z]/(yz-x^2)$$ and the projective parabola $X=\operatorname{Proj}(R)$. Let $I=\langle y,z\rangle\subseteq R$ be the homogeneous ideal generated by $y$ and $z$, and consider the ideal sheaf $\newcommand{\cL}{\mathscr{L}}\cL:=I(1)^\sim$, as in, $\cL(X)=I_1=\Bbbk y \oplus \Bbbk z$. Observe that it is globally generated because $y$ and $z$ do not vanish simultaneously on $X$, otherwise $x$ would have to vanish as well. However, of course, the morphism $X\to\newcommand{\P}{\mathbb P}\P^2$ defined by $[x:y:z]\mapsto [y:z]$ is a finite surjection of degree $2$.

Now, there must be some $k\in\mathbb N$ (probably $k=2$) such that $\cL^{\otimes k}$ is very ample. However, I don't understand how it works, in this very concrete example.

My question is basically: what are the global sections of $\cL^{\otimes k}$ and does $\cL^{\otimes 2}$ contain functions $xz$ and $xy$?

That would be wonderful, but I can't see how to define either of them on both $X_y$ and $X_z$. More precisely, I would think that $$\cL^{\otimes 2}(X_y)=\left\{ \frac{fz^2}{y^n} \:\middle\vert\: f\in R_n \right\}$$ but if $xz=fz^2y^{-n}$ then $x=fzy^{-n}$ and this is hard to do when we only have a relation for $x^2$. I must be mistaken somewhere, and this is driving me crazy.

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Jesko Hüttenhain
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    do you have a precise definition of $\tilde{I(1)}$ somewhere ? shouldn't $x = xy/y = xz/z$ belong to $\cL (X)$ ? Your $X$ is isomorphic to $\Bbb P^1( \Bbb k)$, where the ample line bundles are very ample. – mercio Nov 09 '13 at 17:27
  • I am using Hartshorne's definition, but you are right, the global sections of $I(1)^\sim$ already contain $x$. It was a silly question after all. Still, I am yet to find the example I am looking for. It's fine to know that there are abstract examples, but I would like to see this in coordinates at least once. Sigh. – Jesko Hüttenhain Nov 09 '13 at 17:35
  • @mercio: Post that as an answer. Would be a shame if the 400 rep would just go to waste. – Jesko Hüttenhain Nov 10 '13 at 10:57
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    @JeskoHüttenhain: Hi Jesko, I think the two easiest examples are to take $C$ either an elliptic curve, with $L$ the line bundle of any degree-2 divisor, or to take $C$ a genus-2 (hyperelliptic) curve, with $L$ its canonical bundle. In both cases, $L$ is ample and globally generated, and gives a map to $P^1$ (so it is clearly not very ample). For the elliptic case, $L^{\otimes 2}$ is very ample; for the hyperelliptic case, $L^{\otimes 3}$ is very ample (and $L^{\otimes 2}$ is not).

    Edit: In coordinates, you can find the genus-2 curve as a smooth curve of type (2,3) on $P^1 \times P^1$.

     – Jake Levinson Jan 30 '14 at 04:18
  • @JakeLevinson: Thanks a bunch! – Jesko Hüttenhain Jan 30 '14 at 08:44

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$X$ is isomorphic to $\Bbb P^1(\Bbb k)$,by the maps induced by $(u:v) \mapsto (uv:u^2:v^2)$ and $(x:y:z) \mapsto (y:x) = (x:z)$.

$x$ is a global section of $\cL$, because $x = xy/y \in \cL(X_y)$ and $x=xz/z \in \cL(X_z)$, and those two are equal on $X_y \cap X_z = X_x$ : $xz/z = xyz/yz = xy/y$, so $x \in \cL(X)$. Actually, the pulled back sheaf of $\cL$ to $\Bbb P^1(\Bbb k)$ is $\mathcal O_{\Bbb P^1(\Bbb k)}(2)$, which is ample and also very ample.

mercio
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