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When speaking about blow-ups, I've seen that everybody says "let $Y \subset X$ be a smooth closed subvariety of codimension $r$..."

What happens when one want to blow-up subvarieties which are not of pure dimension?

You can always decompose $Y$ into a union of pure dimension subvarieties and blow-up them successively. Does the order matter? Or the assumption of $Y$ being smooth mean that these components cannot intersect?

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2 Answers 2

Blowing up commutes with passing to open sets in $X$. Therefore, if two components don't intersect the blowups commute. And yes, $Y$ smooth would imply the components are smooth and don't intersect. (According to some authors "smooth" should also imply that $Y$ is of pure dimension!)

Even if $Y$ is of pure dimension, but has intersecting components, the order can matter. Consider the parable of the orthant $\mathbb R_+^3$. If we use a planing device to shave off edge #1, then (less of) what remains of edge #2, we get a combinatorially inequivalent polyhedron than if we do #2 first, then #1. This polytopal picture illustrates (via the toric variety dictionary) what happens if we blow up one coordinate line, then the proper transform of another, in $\mathbb A^3$.

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Note that the blow-up of the union of irreducible is not the same as the consecutive blow-ups of the components (except if they are disjoint, as Allen Knutson explained). For example the blow-up of two lines intersecting in the projective space is singular (check in coordinates) and the two ways of blowing-up the lines one after the other give two smooth threefolds.

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Right, blowing up both together corresponds to planing off those two edges of the orthant the same amount, rather than the second one less than the first. The resulting 3-d polyhedron has a vertex of degree 4. – Allen Knutson Feb 28 '14 at 7:23

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