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?


2 Answers 2


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!

  • $\begingroup$ 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

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 $- 2 \tilde{D}_1$

  • 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

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