I would like some help understanding the proof in $(\impliedby)$ direction.

Hartshorne on page 156, Example 7.6.2 says:

If $\mathcal{L}$ is an invertible sheaf on $\mathbb{P}^1 \times \mathbb{P}^1$ of type $(a,b)$, i.e. $\mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(a,b)$, with either $a<0$ or $b<0$, then by restricting to a fibre of the product $\mathbb{P}^1 \times \mathbb{P}^1$, one sees that $\mathcal{L}$ is not generated by global sections.

i.e. wlog let $a<0$. Then restricting to $\mathbb{P}^1 \times \{x\}$, we get a sheaf $\mathcal{O}_{\mathbb{P}^1}(a)$, which is not generated by global sections, as it doesn't have any when $a<0$.

That means $\mathcal{L}$ can not be very ample (since a very ample sheaf is always generated by finitely many global sections). In particular, by the same reasoning, no positive tensor power of $\mathcal{L}$ can be very ample.

Finally, that implies that $\mathcal{L}$ can not be ample.

If $a=0$, $\mathcal{O}_{\mathbb{P}^1}(a) = \mathcal{O}_{\mathbb{P}^1}$, which is not ample, since by definition (Hartshorne p 153) a sheaf $\mathcal{L}$ is ample if for every coherent sheaf $\mathcal{F}$ there is an integer $n_0>0$ such that for every $n \geq n_0$, the sheaf $\mathcal{F} \otimes \mathcal{L}^n$ is generated by global sections. In our case $\mathcal{L}=\mathcal{O}_{\mathbb{P}^1}$ and take $\mathcal{F}=\mathcal{O}_{\mathbb{P}^1}(-1)$. It is well known that the ideal sheaf $\mathcal{O}_{\mathbb{P}^1}(-1)$ is coherent, so we can use the above definition. So $\mathcal{O}_{\mathbb{P}^1}(-1) \otimes \mathcal{O}_{\mathbb{P}^1}^n$ is just $\mathcal{O}_{\mathbb{P}^1}(-1)$ for any $n>0$, which is not globally generated.

Hence, if $a \leq 0$ or $b \leq 0$, $\mathcal{L}$ can not be ample.

Is the above correct?

I think I'm using that the restriction of a globally generated sheaf to a subvariety is also globally generated. Similarly for very ample sheaves.

  • 2
    $\begingroup$ Your argument is pretty much correct. The only part I don't like is when you say "As for the case when a=0, the global sections of $O_{P^1}$ is a 1-dimensional vector space, thus a rational map to $P^n$ can not be defined." In fact sections of $O_{P^1}$ give a perfectly good map to the projective space $\mathbf P^0$. Instead, I would simply argue that the restriction of an ample line bundle to a subvariety is still ample. Since $O_{P^1}$ isn't ample on $\mathbf P^1$, that proves what you want. $\endgroup$ – user64687 Apr 15 '15 at 10:38
  • $\begingroup$ Thank you, I've erased the wrong part. $\endgroup$ – Kristina Apr 15 '15 at 15:55
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    $\begingroup$ OK, but now there is a little gap, because you don't deal with the case $a=0$. That was what my comment was meant to address. $\endgroup$ – user64687 Apr 15 '15 at 15:57
  • $\begingroup$ In a minute! :) $\endgroup$ – Kristina Apr 15 '15 at 16:00
  • $\begingroup$ OK, now it seems complete! $\endgroup$ – user64687 Apr 15 '15 at 17:40

I can try here but I might be wrong. I just write down some ideas.....maybe someone can come up with a better one.

I think it is easier to look at the following commutative diagram...

$$\require{AMScd} \begin{CD} \text{Spec}(k)\times \mathbb{P}^{1} @>{j}>> \mathbb{P}^{1}\times\mathbb{P}^{1}@>{p}>> \mathbb{P}^{1}\\ @V{f}VV @V{g}VV @V{h}VV\\ \text{Spec}(k) @>{i}>> \mathbb{P}^{1}@>{q}>> \text{Spec}(k) \end{CD}$$ where the two small squares and the huge rectangle are cartesian, and the products are all over $\text{Spec}(k)$.

[I think that is what he means by looking at a certain fiber]

Suppose we have $\mathcal{O}(a)$ on $\mathbb{P}^{1}$ in the (2,2) position at $\mathcal{O}(b)$ on $\mathbb{P}^{1}$ in the (1,3) position, with $b<0$. Hence the corresponding line bundle on $\text{Spec}(k)\times\mathbb{P}^{1}$ is $f^{*}i^{*}\mathcal{O}(a)\otimes j^{*}p^{*}\mathcal{O}(b)\cong j^{*}p^{*}\mathcal{O}(b)$.

So if $\mathcal{L}$ has a global section, then $j^{*}p^{*}\mathcal{O}(b)$ has a global section. But $\text{Spec}(k)\times\mathbb{P}^{1}\cong\mathbb{P}^{1}$, we have

$$\require{AMScd} \begin{CD} \mathbb{P}^{1} @>{p\circ j}>> \mathbb{P}^{1}\\ @V{f}VV @V{h}VV \\ \text{Spec}(k) @>{q\circ i}>> \text{Spec}(k) \end{CD}$$

then I think it can be checked by universal property of cartesian product that $p\circ j$ is identity. Hence $j^{*}p^{*}\mathcal{O}(b)=\mathcal{O}(b)$ will have a global section, which is a contradiction.

  • $\begingroup$ This answer seems very interesting, but I'm afraid I don't understand it. $\endgroup$ – Kristina Apr 15 '15 at 16:13
  • $\begingroup$ No worries. I was in fact saying the same thing as you did but I tried to make it rigorous. Not sure if it succeeds. Cheers! $\endgroup$ – enoughsaid05 Apr 15 '15 at 20:19

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