1
$\begingroup$

I got a contradiction. Help me to understand it. It is about answer by Sándor Kovács of a question https://mathoverflow.net/questions/55526/example-of-a-variety-with-k-x-mathbb-q-cartier-but-not-cartier

Kovács gave an example $X$. Basically this $X$ is a toric variety. It is not hard to calculate class group of $X$. For any toric variety class group is generated by tor-invariant divisors. For $X$ there are three such divisors. Set theoretically they are defined by equations $x^2=0$, $y^2=0$ and $z^2=0$. But those functions have zero of order two on corresponding divisors. So it is better to write

$$D_1 = \frac{1}{2} div( x^2 )$$ $$D_2 = \frac{1}{2} div( y^2 )$$ $$D_3 = \frac{1}{2} div( z^2 )$$

Relations between these divisors are generated by monomials. $$D_1 + D_2 = div (xy) = 0$$ $$2 D_3 = div(z^2) = 0$$ $$ \dots $$

Relations are as follows $D_1 + D_2$, $D_2 + D_3$, $D_3 + D_1$, $2D_1$, $2D_2$, $2D_3$.

Evidently the class group which is generated by $D_1$, $D_2$ and $D_3$ modulo relations is just $\mathbb{Z} / 2 \mathbb{Z}$. After tensoring with $\mathbb{Q}$ it becomes 0.

So the $K_X$ is a $\mathbb{Q}$ divisor which is just zero. But $\pi^* K_X$ is not zero as a $\mathbb{Q}$ divisor (it follows from Kovács calculations).

Question: What is wrong?

$\endgroup$

2 Answers 2

2
$\begingroup$

You are right, there was a tiny mistake in my answer, although it did not change anything. The degree of $V$ is $4$, not $2$, but as you point out the calculation still leads to getting $\dfrac 12 E$.

Furthermore, since you calculated that $K_X$ generates the class group $\mathbb Z/2\mathbb Z$, it shows the original claim that $K_X$ is not Cartier, but $2K_X$ is.

For the record, $\dfrac 12E$ is not integral despite any calculation. You can say that $E$ is linearly equivalent to twice an integral divisor, but that is not the same. One has to be careful distinguishing between divisors and divisor classes!

$\endgroup$
8
  • $\begingroup$ Sorry but your answer is still not correct. $E^2 = - 2 L$. It is a general fact: if you consider zero section $Z$ in total space of a line bundle $\mathcal{O}(D)$ then $Z^2 =D$. $\endgroup$
    – quinque
    Nov 27, 2015 at 11:46
  • $\begingroup$ You are right, it is $-2L$. But that just means that my original calculation was correct. :) $\endgroup$ Nov 30, 2015 at 6:42
  • $\begingroup$ Calculation was correct. But it does not imply that $K_X$ is not Cartier. $\endgroup$
    – quinque
    Nov 30, 2015 at 14:28
  • $\begingroup$ Actually it does. I will add some stuff to the answer so you can see it. $\endgroup$ Dec 2, 2015 at 5:19
  • $\begingroup$ I read your edits. Generally there are some problems with pull back of Weil divisors. I even found your post about it here mathoverflow.net/questions/16888/when-do-divisors-pull-back . Why this problems do not arise in this case? $\endgroup$
    – quinque
    Dec 2, 2015 at 8:32
0
$\begingroup$

Point one. $K_X$ is indeed $\mathbb{Q}$-Cartier but not Cartier. I calculated class group (see my question). It is $ \mathbb{Z} / 2 \mathbb{Z} $. $K_X$ represents the nontrivial element of class group. To see this we will construct an explicit section $\omega$, which equals zero precisely on divisor $D_1$. Pull back of this section on $\mathbb{A}^3$ equals $x dx \wedge dy \wedge dz$. If you want to be sure that this $\mathbb{Z}/ 2 \mathbb{Z}$ invariant form is indeed a form on $X$ you may write it in two chats:

$$\omega = \frac{d(x^2) d(y^2) d(z^2)}{8 yz}, \ \ \text{if} \ \ y \ne 0, \ \ z \ne 0$$

$$\omega = \frac{d(x^2) \wedge d(x y) \wedge d(x z)}{2x^2} \ \ \text{if} \ \ x \ne 0$$

To see that this two section matches on the overlap, one can take pull back on $\mathbb{A^3}$. To verify that zero locus of this section is indeed $D_1$ one should consider pull back on $\mathbb{A}^3$.

Point two The proof by Sándor Kovács is wrong (at least as far as I understand it). He claims that $K_Y = \pi^* K_X + \frac{1}{2} E$. And conclude that $\pi^* K_X = K_Y - \frac{1}{2} E$ is not integral (here is assumed that $\frac{1}{2} E$ is not linear equivalent to integral divisor)

The point is that $\pi^* K_X = 0$. And $\frac{1}{2} E$ is linear equivalent to integral.

Consider $\pi^* x^2 \in k[Y]$ $-$ pull back of $x^2 \in k[X]$.

It indeed has zero of order two on $\tilde{D}_1$ (which is proper transform of $D_1$). But also it has zero of order 1 on exceptional locus. To explain this I remind you that $Y$ is a total space of $\mathcal{O} (-2)$. One may treat $x^2$ as an element of $\Gamma ( \mathbb{P}^2, \mathcal{O}(2) )$. Section of dual bundle $\mathcal{E}^*$ always defines a function on total space of $\mathcal{E}$ which is linear on fibers (in our example $\mathcal{E} = \mathcal{O}(-2)$). I leave it as an exercise to convinse yourself that this function coincides with $p^* x^2$. So $p^* x^2$ has zero of order one on exceptional locus (it is linear function on fibers, so it has zero of order one on zero section).

To conclude

$$div ( p^* x^2 ) = 2 \tilde{D}_1 + E$$

So $E$ is linear equivalent to $- 2 \tilde{D}_1$

$\endgroup$
2
  • 1
    $\begingroup$ Sándor Kovács proved everything correctly. I am sorry for my unfair accusations. He really works with actual divisors (not with divisor classes). I am thankful to Sándor Kovács for explanations. $\endgroup$
    – quinque
    Feb 11, 2016 at 11:45
  • $\begingroup$ My problem was purely psychological. You can do two thing things with group, generated by divisors: 1. quotient by rationality relation 2. tensor by $\mathbb{Q}$. You can do either of this things or both of them. For some reason I thought, that applying only the second thing is odd. But it actually helps in this example. $\endgroup$
    – quinque
    Feb 11, 2016 at 11:50

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.