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Added a comment finding a reference for Zariski factorization.
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Karl Schwede
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.

Edit: I've looked around for a good reference on "Zariski-Factorization", but I'm not sure what a good one is. Does someone know?

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.

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.

Edit: I've looked around for a good reference on "Zariski-Factorization", but I'm not sure what a good one is. Does someone know?

Source Link
Karl Schwede
Karl Schwede
  • 20.5k
  • 3
  • 53
  • 98

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.