5
$\begingroup$

Let $X$ be a normal variety over $\mathbb{C}$.

In their book Birational geometry of algebraic varieties, Kollár and Mori define [Definition 2.50 and 2.51] a ramified m-th cyclic cover associate to a line bundle $L$ ramified along $D \subseteq |mL|$ to be the relative spec $$Spec_X(\oplus_{i=0}^{m-1}L^{-i}).$$ Or more generally, for a rank $1$ torison free sheaf $L$, the m-th cyclic ramified cover is $$Spec_X(\oplus_{i=0}^{m-1}L^{[-i]}),$$ where $L^{[i]}$ is the double dual of $L^{\otimes{i}}$.

In the book Singularities of the Minimal Model Program, Kollár and Kovács give a definition of ramified cover [see Definition 2.39], which is roughly as follows:

A finite morphism of normal schemes $\pi: \tilde{X} \to X$ is called a ramified cover of degree m if there is a dense open subset $U \subseteq X$ that contains every codimension $1$ point of Sing$X$ such that the restriction $\pi_U: \tilde{U}\to U$ is etale and has degree m.

My question: Is the (general) ramified m-th cyclic cover in the sense of Kollár and Mori a special case of the ramified cover in the sense of Kollár and Kovács?

In my case, $X$ has canonical singularities, and $D$ can contain the codimension 1 point of Sing$X$. So, when choose $U$, it is inevitable to intersect $D$, hence I worry if the resulting morphism etale?

My interest in the problem is because I want to know the singularity of the ramified cyclic cover $\tilde{X}$ (in the sense of Kollár and Mori). Again, $X$ has canonical singularities, I want to know if $\tilde{X}$ has the same singularities.

By the book of Kollár and Kovács (See Page 65-65), it claims that the discrepancy does not get worse by taking a finite ramified cover (in their definition). I looked at the proof, and feel it could go through without any change for the (general ) cyclic ramified cover case. Did I miss something?

Improve this question
$\endgroup$
10
  • 2
    $\begingroup$ Be careful, because the cyclic covers as defined need not be normal if $D$ is non-reduced: in the simple example $X=\mathbb{A}^1$, $D=2\cdot (x)$ the recipe gives the non-normal $\tilde X = {\rm Spec}(k[x,y]/(x^2-y^3)$. But it's easy to compute the normalization, which is ${\rm Spec_X} \bigoplus_{i=0}^{n-1} \mathcal{O}_X(\lfloor \frac{i}{n}D \rfloor)$. $\endgroup$
    – Piotr Achinger
    Commented Feb 9, 2015 at 17:45
  • 1
    $\begingroup$ In general the cyclic cover will be ramified everywhere on $D$, so if $D\cap {\rm Sing}(X)$ is big there is no hope. $\endgroup$
    – Piotr Achinger
    Commented Feb 9, 2015 at 17:48
  • $\begingroup$ And of course, cyclic covers in characteristic $p$ can be inseparable (and hence not etale). Also, the discrepancy arguments can be different in characteristic $p$ even if the cover is separable (due to wild ramification). But outside of the cases already mentioned in the comments: yes every cyclic cover is a a ramified cover in the sense of Kollár and Kovács. $\endgroup$
    – Karl Schwede
    Commented Feb 9, 2015 at 18:13
  • 1
    $\begingroup$ Maybe I should say Kollár and Kovács really want to treat carefully the non-normal case (because they really need to handle that case in moduli applications). $\endgroup$
    – Karl Schwede
    Commented Feb 9, 2015 at 22:38
  • 1
    $\begingroup$ I strongly believe that Kollár-Kovács mean "the points of $\text{Sing} X$ that are codimension 1 in $X$", so there is nothing to worry about. The condition of being codimension $1$ in $\text{Sing} X$ is not a relevant condition to the sorts of questions they are studying. $\endgroup$
    – Karl Schwede
    Commented Feb 9, 2015 at 23:42

1 Answer 1

Reset to default
5
$\begingroup$

Ok, so based on the discussion in the comments, maybe I should put this into an answer. I think the confusion comes from the phrase

every codimension 1 point of $\text{Sing }X$.

What the authors Kollár and Kovács mean here is to consider

every point of $\text{Sing }X$ that is also a codimension $1$ point of $X$.

I agree that this could be interpreted in other ways, but this is what the authors mean (I'm sure Sándor will agree if he sees this).

Given this, and that $X$ is normal over $\mathbb{C}$ (or any field of characteristic zero), it is easy to see that every cyclic cover is a ramified cover in the sense of Kollár-Kovács (as I think you already see). As pointed out in the comments, if $X$ is non-normal, or if we are working in characteristic $p > 0$, life becomes more complicated.

Your particular situation

You had $X$ with canonical singularities and $\widetilde{X}$ a ramified cyclic cover. Then it is easy to see that $(X, -\text{(Ramification Divisor)})$ also has canonical singularities (notice we have a non-effective divisor here). For a proof simply see Kollár-Mori 5.20(3).

Improve this answer
$\endgroup$
1
  • $\begingroup$ I can confirm that Karl is completely right: 1) we meant the codimension $1$ points of $X$ that lie in $\mathrm{Sing}X$ and 2) our main concern was to allow covers of non-normal schemes. An interesting point that shows why one needs to exclude codimension $1$ points is that for instance the Hurwitz formula (see page 65) fails for the normalization map. More generally, the normalization map works sort of the opposite way than ramified covers (see what I mean in (5.7) on page 191) and then one runs into difficulty comparing discrepancies.... $\endgroup$
    – Sándor Kovács
    Commented Feb 10, 2015 at 22:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .