Embedded resolution of singularities - MathOverflow

Embedded resolution of singularities

2013-02-01T15:18:19Z

I'd like to check with my colleagues whether I have correctly understood "embedded resolution of singularities".

Let $X$ be a nonsingular projective variety over $\mathbf C$ and let $D$ be a "nice" divisor on $X$, say $D$ has strictly normal crossings. (Maybe we could just take $D$ to be a closed subscheme?)

Then, has the following statement been proven? And what is a "good" reference?

There exists a projective birational surjective morphism $\psi:Y\to X$ with $Y$ a nonsingular projective variety over $\mathbf C$ and the inverse image of $D$ in $Y$ a nonsingular projective variety over $\mathbf C$ of codimension one in $Y$?

I'm worried about whether I have correctly understood this statement, or maybe one needs some "normality" conditions on $D$ to assure this "embedded" resolution of singularities.

Also, how does one obtain this embedded resolution of singularities? Can we write down a terminating process which ends with an embedded resolution of singularities?

I have a hard time "believing" the above statement, but I don't know why. If anybody can explain to me that this is not so surprising as a result I would be very thankful.

Answer by Donu Arapura for Embedded resolution of singularities

2013-02-01T21:54:02Z

What embedded resolution does achieve is this: if $Z\subset X$ is a Zariski closed set in a variety, then there is a smooth variety $Y\to X$, obtained from a series of blow ups along smooth centers, such that the preimage of $Z$ is a divisor with normal crossings. However, there is no reason to expect that the inverse of image of $Z$ (or your $D$) in a further blow up will be smooth. (Francesco said this already in a comment, but it bears repeating.) This is already clear in the case where D is a union of two lines in the plane. If you take the preimage in a further blow up, you will get a tree of $\mathbb{P}^1$'s. So your instinct is correct here.

Answer by Sándor Kovács for Embedded resolution of singularities

2013-02-02T00:15:51Z

As Donu and Francesco already explained, the preimage of $D$ is usually not smooth simply because if it contains any subvarieties (subschemes) that are blown up, then the preimage will not be irreducible, but if $X$ is normal, then the preimage will be connected and hence necessarily singular along the intersection of different components.

On the other hand what can be achieved is that the strict transform of $D$ is non-singular. The strict transform or birational transform is the closure of the dense subset of $D$ that lies on the locus of $X$ where the resolution is an isomorphism.