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If $X$ is a normal projective variety and $D$ a divisor in it, we say that $\pi\colon (\widetilde X,\widetilde D)\rightarrow (X,D)$ is a log resolution if $\widetilde X$ is a resolution of $X$, the strict transfor $\widetilde D$ of $D$ is non-singular and $\widetilde D \cup \{ \text{exceptional divisors} \}$ have simple normal crossings.

I know little about techniques for resolution of singularities and as far as I am aware, for varieties over algebraically closed fields, the problem of finding resolutions of singularities is open.

However, I was wondering if the following is solved, by whom and if someone can provide me with a 'black-box' reference:

Question: Given a projective variety $X$ over a field of characteristic $p$ and a divisor $D$ on $X$, is there a log resolution of the pair $(X,D)$ in the cases where $X$ is a non-singular variety of small dimension (1,2,3) and/or in the case the $p\neq 2,3,5\ldots$? What if the variety is a product of a projective variety and the affine line?

Of course partial answers are appreciated. However the purpose of this is just to use it in a birational proof for something else, so by no means I intend to prove it myself or get any close to it.

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See Resolution of singularities for varieties in dimensions 1,2,3 have been proved in all characteristics. – Parsa Oct 24 '11 at 15:02
That is what made me post the question. I would like to have an actual reference other than Wikipedia. Moreover WP speaks just of resolution of singularities. What I am aiming for is \em{log resolutions} of pairs $(X,D)$. Is it 'obvious' that if $X$ has dimension at most $3$ and therefore $D$ has dimension at most $2$ I can find a resolution $\pi\colon(\widetilde X,\widetilde D)\rightarrow(X,D)$ such that $\widetilde X$ is a resolution of $X$ and $\widetilde D$ is non-singular \em{and $\pi^{-1}(D)$ consists of simple normal crossings} just because I can resolve $X$ and $D$ independently? – Jesus Martinez Garcia Oct 24 '11 at 21:12
I don't know if embedded resolution of singularities is known in dim 3. Without the $D$ and $p> 5$ (for some weird technical reason) it goes back to Abhyankar in the 60's, who also did the dim. 2 case earlier. There are newer proofs, I believe in unrestricted char and dim 3, due to Cutkosky Amer J 2009, and Cossart-Piltant J Algebra 2008/9. The link that Parsa gives more precise refs. – Donu Arapura Oct 24 '11 at 22:49
up vote 3 down vote accepted

What you want does follow from Cutkosky's paper cited by Donu Arapura (and presumably already from results of Abhyankar, but I have not checked). One just has to combine his Theorems 1.1 and 1.2. More precisely, one can resolve singularities of $X$ using Theorem 1.1 to get $\pi_1: X_1 \to X$ with $X_1$ smooth and then apply Theorem 1.2 to the pair $(V,S) = (X_1,D_1)$, where $D_1$ is the union of $\pi_1^{-1}(D)$ and the exceptional divisors.

This requires that the characteristic be $>5$, but if you use the result of Cossart and Piltant instead of Cutkosky's Theorem 1.1 you can get a log resolution in any characteristic (since Cutkosky's Theorem 1.2 has no characteristic restriction).

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You are right. The same is explained by Hauser in… It requieres a bit more than embedded resolutions because one wants to have log resolutions but Hauser explains this too. Thanks :) – Jesus Martinez Garcia Oct 25 '11 at 10:19
However note that Cutkowsky, Cossart and Piltant paper are for non-singular threefolds! The case of $(X,D)$ where $X$ is a singular threefold is open. – Jesus Martinez Garcia Oct 25 '11 at 11:21
As I explained, one can first resolve the singularities of $X$ to reduce to the nonsingular case. The only potential problem I can see is that the exceptional locus or the inverse image of $D$ might not be pure dimensional. Is that what you are worried about? (It seems to me that this can be fixed by applying Cutkosky's Theorem 1.3.) – ulrich Oct 25 '11 at 12:41
If you are referring to Hauser's comment that embedded resolution is open for threefolds, I think what he means is that it is not known when $D$ has dimension $3$, not $X$. – ulrich Oct 25 '11 at 12:42
First: it is enough to get embedded resolution, since one can get a log resolution from it following footnote 4 in Hauser. I am not an expert so for me your comment on reducing to non-singular case seems reasonable. Even if the inverse image of $D$ was not pure dimensional, one could do the modification above to make it so. However the strict transform of $D$ must also be pure dimensional and I am not sure this can be achieved in that way. – Jesus Martinez Garcia Nov 2 '11 at 12:32

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