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Let $X$ be a smooth and projective variety over a field of characteristic zero. Let $Y$ be a normal variety, with finite quotient singularities (an orbifold!) and let $\pi: Y \to X$ be a finite morphism, ramified along a simple normal crossings divisor $D$. Assume that the singularities of $Y$ lie over the singular locus $D_{sing}$, i.e. $\pi(Y_{sing}) \subset D_{sing}$.

Now $Z \subset X$ is a smooth closed subvariety of codimension 2 in $X$, intersecting transversally all $D_J:=\bigcap_{i \in J} D_i$ (here $D=\sum_{i \in I} D_i$, with $D_i$ the irreducible components).

Consider the blow-up $\tilde{X} \to X$ of $X$ along $Z$ and form the cartesian product $\tilde{Y}=Y \times_X \tilde{X} \longrightarrow \tilde{X}$.

My question is: is $\tilde{Y}$ still normal and with finite quotient singularities?

Thanks for your help!

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I'm confused, is $Z$ smooth itself? Otherwise I don't think this can be true. Blowups aren't necessarily normal or have quotient simngularities. If $\pi$ is an isomorphism then $\tilde{X}$ isn't necessarily normal and doesn't necessarily have finite quotient singularities. $$\text{ }$$ Also, is $D$ a divisor on $X$ or on $Y$? –  Karl Schwede May 24 '13 at 12:51
    
Karl, thanks for your answer. I was assuming $Z$ is smooth (I forgot to write it), so there is no problem with $\tilde{X}$: it is also smooth. And the divisor is on $X$: I wanted to say that $\pi(Y_{sing}) \subset D_{sing}$. –  bloworbifold May 24 '13 at 13:34
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2 Answers

up vote 1 down vote accepted

I think this is true. Taking a point $x\in X$, and base change your setting up to its formal neighborhood, we can assume $K(X)\subset K(Y)$ is a Galois extension with finite abelian group.

Now we can put a root stack structure $\mathcal{X}\to X$ which is branched over $D$ such that the morphism $Y\setminus\pi^{-1}(V)\to \mathcal{X}\setminus{V}$ is etale where $V\subset \mathcal{X}$ is of codimension at least 2 (See Matsuki-Olsson's paper). Then by purity, this indeed gives a DM-stack $\mathcal{Y}$ which is finite etale over $\mathcal{X}$. In particular $\mathcal{Y}$ is smooth and it has Y as its coarse moduli space. (By the way, I think once you assume $Y\to X$ branches over a snc divisor and $Y$ is normal, it comes for free that $Y$ has only abelian quotient singularities.)

Now the condition you have just implies the preimage $\mathcal{Z}$ of $Z$ in $\mathcal{X}$ is smooth. So its preimage $\mathcal{Z}_{\mathcal{Y}}$ in $\mathcal{Y}$ is smooth.

But then $Bl_{\mathcal{Z}_{\mathcal{Y}}}\mathcal{Y}$ is smooth, so its coarse moduli space only has abelian quotient singularities, which is precisely $\tilde{Y}$.

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Thanks for the answer! –  bloworbifold May 26 '13 at 17:04
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I don't see how $D$ makes any difference in the question you describe. Perhaps you forgot a condition about $D$ such as $\pi$ is ramified along $D$?

In any case, here is an example that shows that you need more conditions to have a chance here.

Let $Y=X=\mathbb A^2$ and $\pi: \mathbb A^2\to \mathbb A^2$ be defined by $(x,y)\mapsto (x^2,y^2)$. Let $Z\subset X$ be the origin. Then $\widetilde X$ is just $\mathbb A^2$ blown-up at a closed point (in particular smooth, normal, etc.), but $\widetilde Y$ is the blow-up of $\mathbb A^2$ along the fat point defined by the ideal $(x^2,y^2)$ so it is a pinch point and hence not normal and not even normal crossings.

I guess now you could say that you want $\pi$ ramified along a divisor, but it will be the same thing, if $Z$ is a point on $D$, it's pre-image will be a fat point and blowing that up will not be normal. It seems that the only chance is to require that $Z$ (or at least its generic point) is disjoint from the ramification locus of $\pi$, in other words if $\pi$ is étale in a neighbourhood of (the generic point of) $Z$.

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Hi Sandors, thanks for the answer. Yes, $D$ is the ramification locus of $\pi$. I cannot ask $Z$ to be disjoint from $D$ for the applications I have in mind, but $Z$ intersect transversally all the $D_J:=\bigcap_{i \in J} D_i$ where $D_i$ are the irreducible components of $D$ and $J$ is any subset of the set of indexes $I$. Is that enough? –  bloworbifold May 24 '13 at 16:18
    
I edited the question –  bloworbifold May 24 '13 at 16:28
    
CX's answer answers this, so I don't need to... :) –  Sándor Kovács May 25 '13 at 1:01
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