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Karl Schwede
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Although there is a good reason that $(x,y)^2$ has a smooth blow-up. It is a power of an ideal which itself has a smooth blowup. See for example Hartshorne, Algebraic Geometry, Chapter II, Section 7, Exercise 7.11.

I suspect that, on smooth surfaces, one can probably say more, via "Zariski-factorization" type ideas, but I'm not sure what the right answer would be.

Karl Schwede
  • 20.5k
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  • 53
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