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Is there a relative version of resolution of singularities (in characteristic 0)?

For finite type morphism $f:X\rightarrow S$ where $S$ is a variety over a field $k$ with char $k=0$(or more generally excellent scheme with residue field are characteristic 0?). Are there modifications $i:S' \rightarrow S$, $j:X'\rightarrow X$, and morphism $f':X'\rightarrow S'$ s.t. $fj=if'$, and $f'$ smooth, $i$ proper, surjective. I'm not sure the above statement is a proper version of relative desingularities.

Thanks

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If $S=spec(k)$, $X$ is a singular variety, I don't think you can use base change to resolve that. –  Honglu Apr 9 '13 at 2:45
    
For finite morphisms, these kind of statements are called simultaneous resolution. But mostly there are negative results. –  Manish Kumar Apr 9 '13 at 3:22
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Even combining pullback and desingularization, one can't resolve the singularities in the Legendre family of elliptic curves $y^2=x(x-1)(x-\lambda)$, because the unipotent local monodromy in $H^1$ won't go away. –  Will Sawin Apr 9 '13 at 3:30
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Perhaps you should look at "semistable reduction" of Kempf, Knudsen, Mumford and Saint-Donat. I believe this is the best you can hope for without some additional hypotheses. –  Jason Starr Apr 9 '13 at 10:43
    
For families of curves, you might find the following article interesting: B. Teissier, "Résolution simultanée I: famille de courbes" (available on Numdam). There is also "Résolution simultanée II" in "Séminaire sur les singularités des surfaces" (1976) but I don't know whether it is available in digital form. –  Damian Rössler Apr 9 '13 at 11:45

1 Answer 1

There are a bunch of very basic fundamental problems with doing this.

Here is one example and you are invited to generalize this to get others: Let $X$ be a smooth surface and $f$ a surjective flat morphism to a smooth curve $S$. If $f$ is not already smooth, then it is not possible to make it smooth the way you suggest (or any reasonable way).

This is one reason why moduli spaces of smooth objects are in general not compact or put it in another way why compactifying moduli spaces is a non-trivial question. (See $\overline {\mathfrak M}_g$ for instance).

To see that the above claim is true consider the following:

  1. If $f$ has a singular fiber, then for any $i:S'\to S$ the base change morphism: $f_{S'}:S'\times_S X\to S'$ will also have singular fibers.
  2. For any $f':X'\to X$ as in the question, the $fj=if'$ assumption means that $f'$ factors through $f_{S'}$, so in other words you want a $g:X'\to S'\times_S X$ modification such that $f'=f_{S'}g$ plus the rest of the requirements.
  3. $S'\times_S X$ is normal. This is because the fibers of $f$ are CM and hence so are the fibers of $f_{S'}$. Hence $S'\times_S X$ is $S_2$ and by a similar argument it is also $R_1$, so by Serre's criterion it is normal.
  4. There is no such $g$. If indeed $f$ had a singular fiber, then the same singular fiber appears as a fiber of $f_{S'}$. Being a curve its unique resolution is its normalization, so a $g$ like that would have to be a finite morphism of degree $1$. However, since $S'\times_S X$ is normal, Zariski's Main Theorem implies that $g$ would have to be the identity. (For simplicity assume that there is at least one non-unibranched singularity on one of the fibers).
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