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I spent some time looking for an example of non finitely generate graded ring $R=\oplus H^0(X,mD)$ where $D$ is a divisor on a variety $X$ of dimensison $>2$. I know there are several such examples (e.g Zariski's), but they are all on surfaces. I believe there must exist many examples. Do you know any?

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up vote 6 down vote accepted

Here is a way of generating lots of examples:

Start with a variety $X$ with an effective cone which is not rational polyhedral (i.e., not finitely generated) and let $L_1,\ldots,L_r$ be a collection of line bundles on $X$ such that their span $\{L_1^{a_1}\otimes \cdots\otimes L_r^{a_r} | a_1,\ldots,a_r \ge 0\}\,\,$ includes the effective cone. For example, one could take $X$ to be the blow-up of projective space at sufficiently many points, or a K3 surface of maximal Picard number.

Now consider the variety $Y=\mathbb{P}(E)$ where $E=L_1\oplus \cdots \oplus L_r$ and the line bundle $O(1)$ on $Y$. We have for $n\geq 0$, $$ H^0(Y,O(n))\cong H^0(X,Sym^n(E))=\bigoplus_{a_1+\ldots +a_r=n}H^0(X,L_1^{a_1}\otimes \cdots\otimes L_r^{a_r}) $$so that $R(Y,O(1))$ is isomorphic to the sum of all sections of all effective line bundles on $X$: this is usually called the Cox ring of $X$. When the effective cone of $X$ is non-rational polyhedral it is clear that this ring is infinitely generated, since it requires sections from all effective divisor classes.

It is usually a difficult problem in birational geometry to decide when a nef and big divisor is semiample, i.e., some multiple of $D$ is base-point free. A well-known theorem of Zariski says that if $X$ is normal and projective, then $D$ is semiample if and only if the section ring $R(X,D)$ is finitely generated. So this theorem gives a way of producing nef and big divisors which are not semiample. In particular, choosing $X$ such that the nef-big cone is not rational polyhedral, $L=O(1)$ is nef and big, but not semiample, since $R(Y,L)$ is not finitely generated.

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Hi John, nice example, but I think that you need to be a little more specific on the choice of line bundles $L_1,\ldots,L_r$: if they happen to be semi-ample, then the line bundle $\mathcal{O}(1)$ will be finitely generated. In your sum, only non-negative multiples of the lines bundles appear. Presumably it suffices to choose the line bundles $L_1,\ldots,L_r$ so that the intersection of the non-negative span of them with the effective cone is not finitely generated (and there is no need for them to be independent or to span the whole Picard group). –  damiano Feb 24 '11 at 10:17
    
Thanks Damiano, you are absolutely right. –  J.C. Ottem Feb 24 '11 at 10:19
    
Thank you!! It helped a lot. –  Xingle Feb 26 '11 at 0:12
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In their paper "Monge-Ampère equations in big cohomology classes", Boucksom, Eyssidieux, Guedj and Zeriahi give an example (Ex 5.4 page 46 here : http://arxiv.org/abs/0812.3674) of a nef and big line bundle over a smooth projective 3-fold which is not semi-ample. More precisely, every positive current in its cohomology class has poles along some subvariety.

Furthermore, it is well-known (Lazarsfeld, PAG e.g) that a nef and big line bundle has a finitely generated sections ring iff it is semi-ample.

In one word, their construction consists in using the famous example of Serre (and studied by Demailly-Peternell-Schneider) of a flat rank 2 vector bundle $E$ on some elliptic curve $C$, and considering on $V:=\mathbb P(E\oplus A)$ (for $A$ ample on $C$) the tautological line bundle $\mathcal O_{\mathbb P(V)}(1)$.

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In Section 7 of this paper the authors gave an example of a smooth 3-fold $X$ with a divisor $D$ such that:

$\lim_{n \rightarrow \infty} \frac{h^0(\mathcal O_X(nD))}{n^3} = 6+ \frac{2\sqrt 3}{9}$

I would love to here about more natural examples!

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Since the canonical ring is finitely generated, the "most natural" candidate line bundle with such a property would be an anticanonical bundle. Unfortunately, I cannot think of an example at the moment, though... –  damiano Feb 24 '11 at 10:31
    
Dear damiano: thank you for your insight. When you think of such example, please give an answer. –  Hailong Dao Feb 24 '11 at 17:45
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