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Suppose I have a smooth algebraic surface $X$ and a subscheme $Z$ such that the blowup $$Y = Bl(X,Z)$$ is smooth.

Now $Z$ does not have to be smooth, say if it is given by some power of the maximal ideal at a point, but is it nevertheless the case that $Y$ is isomorphic to the iterated blowup of $X$ along smooth points?

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Yes. Every proper, birational morphism between smooth algebraic surfaces can be obtained by the iterated blowup at reduced points. I don't have a reference at hand, but it's contained somewhere in Beauville's book. It's probably also in Hartshorne Chapter V.

The basic idea is that given such a morphism $Y \rightarrow X$, there are finitely many points of $X$ where it is not an isomorphism. You can blow up at these and $Y$ will factor through the blow-up. You repeat this process finitely many times and you'll get an isomorphism.

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The OP is assuming the blowup is smooth already. In which case maybe the statement follows from the cone theorem in MMP? (overkill for surfaces!)

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is it clear that contractions of extremal rays still map to X? Or can we run some relative MMP for Y over X? – user1640 Sep 2 '11 at 0:38
Yes, one can run a relative MMP to get the claim. But, as martin says, this is indeed overkill. – ulrich Sep 2 '11 at 9:52

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