Suppose $X/\mathbb C$ is a smooth projective variety of dimension $d>1$, and $D \subset X$ is a simple normal crossing divisor. For any $k>0$, can I construct a ramified $K$-th cover $f: \tilde X \to X$ for some $k \mid K$, such that $f$ only ramifies over $D$, and $\tilde X$ is still a smooth variety?

This is impossible for curves. For example, if $D$ is a single point, and $k=2$, by Hurwitz formula, the degree of ramification locus $R$ is divisible by $2$. But if the ramification happens on a single point, $\deg R = e_D -1$, where $e_D$ is the ramification index at $D$ and it is an even number, this is a contradiction. But I feel this is a special phenomenon on curves.

A possible candidate of such construction is given by Kawamata's paper "Characterization of abelian varieties" Theorem 17. The construction is roughly as the following (suppose $D$ is just a prime divisor for simplicity).

Let $M$ be an ample divisor and $KM-D$ is very ample. For $1 \leq i \leq d$, choose general $H_i \in |KM - D|$ such that $\sum_i^d H_i +D$ is snc. Let $\phi_i$ be a local equation of $H_i +D$, and set $L=\mathbb C(X)(\sqrt[m]{\phi_1}, \ldots, \sqrt[m]{\phi_d})$. Let $\tilde X$ be the normalization of $X$ in $L$.

I guess the idea behind this construction is that because $$\cap_i^d(H_i+D) = D \cup\{\text{finite points}\}.$$ Ignoring those finite points first, for any point $x \not\in D$, there exists a $\phi_j$ such that $\phi_j(x)\neq 0$ and thus $x$ should not be a ramification point. Then, after taking the normalization, those finite points cannot be ramification points either.

However, I have difficulty to follow Kawamata's construction (I try to write the simple case $\mathbb C[x,y]$ and $D=0$). For me, the above covering have degree $m^d$ (it seems to be a repetition of $m$-th cyclic cover $d$-times), but Kawamata claims that it is of degree $m$ (see the last four line in his proof).

I am very appreciate if one could answer my question or even explain Kawamata's construction to me (even if it may have other ramification locus)!!

  • 4
    $\begingroup$ This is too much to hope for, for example there are no coverings of $X=\mathbf{P}^n$ ramified only along $D= \mathbf{P}^{n-1}$, as $X\setminus D = \mathbf{A}^n$ is simply connected. $\endgroup$ – Piotr Achinger May 23 '18 at 9:19
  • $\begingroup$ Why this is a contradiction? Could it happen that after restricting to $\mathbf A^n$, it is just a trivial etale cover? $\endgroup$ – Li Yutong May 23 '18 at 9:45
  • $\begingroup$ @LiYutong So, what is your definition of a cover? $\endgroup$ – Laurent Moret-Bailly May 23 '18 at 10:01
  • $\begingroup$ I definite it to be a finite morphism and I require $\tilde X, X$ are both irreducible and smooth here. $\endgroup$ – Li Yutong May 23 '18 at 10:19
  • 3
    $\begingroup$ Then Piotr is right. If $\tilde{X}$ is irreducible, the covering cannot become trivial on a (Zariski) open subset. $\endgroup$ – abx May 23 '18 at 10:32

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