Consider the following setting: suppose that $X$ is a smooth variety and let $f:X\rightarrow \Delta$ be a smooth morphism outside the origin $0$. Let the central fiber $X_0$ be a reduced (Cartier) divisor with simple normal crossings. Here a reduced divisor is a Weil divisor $D=\sum_{i}n_i D_i$ with all $n_i=1$. Take some successive blowing ups along the singular locus of $X_0=\sum _{i\in I}X_i$ to get a proper modification $\mu:V\rightarrow X$ from a smooth variety $V$ to $X$. We then write the total transform$\mu^*{X_0}=\sum _{i\in I}\mu_*^{-1}X_i+\sum_{j\in J}k_j V_j^\prime$ with $k_j\geq 1$. Here, we can ensure the strict transform $\sum _{i\in I}\mu_*^{-1}X_i$ to be smooth.

Here is my question, can we determine the ramification divisor of the morphism $\mu$? Namely, if we write $$K_V\equiv\mu^*K_X+\text{Ram}(\mu),$$ then what is $\text{Ram}(\mu)$, should it be written as $t_j V^\prime_j$? If so, what is the relationship between $t_j$ and $k_j$?

  • $\begingroup$ Could you please clarify what you mean by "reduced divisor"? Is this a Cartier divisor (so that the morphism is flat)? If not, could you at least clarify whether or not $X$ is assumed to be normal? $\endgroup$ Jan 2, 2023 at 12:35
  • $\begingroup$ Dear @JasonStarr, thanks for your reply and sorry for my possible misleading. Here a reduced divisor is a Weil divisor $D=\sum_{i}n_i D_i$ with all $n_i=1$. Our $X_0$ is also a Cartier divisor since the defining equation is just $f$ (notice that $X_0=f^{-1}(0)$). And our $X$ is smooth, all fibers except $X_0$ are smooth, and we can use $\mu$such that the strict transform of $X_0$ to be smooth. $\endgroup$
    – Invariance
    Jan 2, 2023 at 14:11
  • $\begingroup$ The pair $(X, X_0)$ is log smooth, but I want to use $\mu$ to make $X_0$ to be disjoint. $\endgroup$
    – Invariance
    Jan 2, 2023 at 14:25
  • $\begingroup$ Is $\sum_{i \in I} \mu_*^{-1} X_i$ really the strict transform? That seems more like the total transform, I think? $\endgroup$ Jan 2, 2023 at 20:22
  • $\begingroup$ If your goal is to make the components of $X_0$ disjoint then you don't need to deal with an arbitrary sequence of blowups. You can just take the intersection lattice of the divisor arrangement (the collection of intersections of subsets of the components of $X_0$, partially ordered by inclusion) and blow them up in order of increasing dimension. Since you assume that you're starting with SNC, this will work. You get one (new) exceptional component for each blowup coming from an intersection; its coefficient is the number of components of $X_0$ that contained that intersection. $\endgroup$ Jan 2, 2023 at 20:28

1 Answer 1


Ok, here's an expanded version of what I said in the comment.

SNC case

First, I'm going to create a special log resolution of the pair $(X, X_0)$. We assume $d = \dim X$. We are assuming that $X$ is nonsingular and $X_0$ is simple normal crossings.

We begin by letting $S_{n}$ denote the stratum of this pair of dimension $n$. In other words, $S_0$ is the points that are intersections of components of $X_0$, $S_1$ is the curves that are components of $X_0$, etc. Let $Y^0 \to X$ be th blowup of all the points in $S_0$. This separates the strict transforms of the components of $S_1$. We let $S_{1}^0$ denote the set of those strict transforms (likewise with $S_{2}^0$, etc). Let $E_{0,i}^0 \subseteq Y_0$ denote the set of exceptional divisors.

Now, we can write $$ K_{Y^0} = \pi_0^* K_X + (d-1) \sum E_{0,i}^0 $$ since we are blowing up nonsingular centers of codimension $d$ in a nonsingular variety. We also ahve $$ \pi_0^* X_0 = (\pi_0)^{-1}_* X_0 + d \sum E_{0,i}^0. $$ since each $E_{0,i}^0$ maps to a point in $S_0$ which is the intersection of $d$ components of $X_0$.

Next, we blowup the curves in $S_1^0$ (these are curves in $Y^0$). Note, each curve may intersect some $E_{0,i}^0$, but it cannot contain any, and likewise no $E_{0,i}^1$ contains any curve in $S_1^0$.. Let $E_{1, i}^1$ denote the exceptional divisors of this new blowup and $E_{0,i}^1$ the strict transforms of the old exceptional divisors. In this case we have $$ K_{Y^1} = \pi_{1,0}^* K_{Y^0} + (d-2) \sum {E_{1,i}^1} = \pi_1^* K_X + (d-1) \sum E_{0,i}^1 + (d-2) \sum E_{1,i}^1. $$ Here we crucially used the fact that no element of the the $S_1^0$ were contained in $E_{0,i}^0$. By a similar consideration, we also have $$ \pi_1^* X_0 = (\pi_1)^{-1}_* X_0 + d \sum E_{0,i}^1 + (d-1) E_{1,i}^1. $$

Continuing in this way, we eventually obtain a log resolution that separates the components of $X_0$, $\pi : Y \to X$ and $$ K_{Y} = \pi^* K_X + (d-1) \sum E_{0,i} + \dots + 1 \sum E_{d-2,i} $$ where the $E_{0,i}$ are the strict transforms of the exceptional divisors from $Y^0 \to X$, $E_{1,i}$ are the strict transforms of the exceptional divisors from $Y^1 \to Y^0$, etc. and also $$ \pi^* X_0 = (\pi)^{-1}_* X_0 + d \sum E_{0,i} + (d-1) \sum E_{1,i} + \dots + 2 \sum E_{d-2, i} $$ In other words for each $E$ an excpetional divisor on $Y$ (or an element of the strict transform), the coefficient of $\pi^* X_0$ is exactly one bigger than the coefficient in the relative canonical divisor (what you called Ram).

SLC case

Let's now assume that $X_0$ is semi-log canonical (SLC) but not necessarily SNC. In this case, since $X_0$ is Gorenstein, this is equivalent to it being Du Bois (see work of Kovács and Doherty). It is also equivalent to the pair $(X, X_0)$ being log canonical by inversion of adjunction.

Anyways, since $(X, X_0)$ is log canonical, for any resolution of singularities, $\pi : V \to X$, if we write $$ K_V + D = \pi^* (K_X + X_0) $$ we know that all the coefficients of $D$ are $\leq 1$ (this is the definition of log canonical). In other words, if we write $$ K_V = \pi^* K_X + \mathrm{Ram}\;\;\;\; \text{ and }\;\;\;\; \pi^* X_0 = (\pi)^{-1}_* X_0 + \sum b_i E_i $$ where $E_i$ are the exceptional divisors (note $\mathrm{Ram} = \sum c_i E_i$ for some $c_i$) then we have that $$ D = \pi^* K_X + \pi^* X_0 - K_V = \pi^* K_X + \pi^* X_0 - \pi^* K_X - \mathrm{Ram} = (\pi)^{-1}_* X_0 + \sum b_i E_i - sum c_i E_i. $$ In other words, for each exceptional divisor $E_i$, we have that the coefficient of $$ D = \pi^* X_0 - \mathrm{Ram} $$ is $b_i - c_i$ and $$ b_i - c_i \leq 1 $$ or in other words $b_i \leq c_i + 1$.

Note, we had equality in the SNC case when I picked a very special resolution of singularities.

  • $\begingroup$ Excellent answer! Thanks a lot! $\endgroup$
    – Invariance
    Jan 11, 2023 at 12:56

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