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Recall that a ring homomorphism A->B is geometrically regular if for all primes p of A, the fiber of B over p is geometrically regular over k(p). A Grothendieck ring (or, G-ring) is one for which A_p->A_p* is regular for all primes p. These are the maps from the local rings of A to their completions.

If A is an order in a number field, is A a G-ring? Equivalently (in this special case), is A excellent?

I've heard it said that 'all' rings that appear in algebraic geometry are excellent. Since an order A corresponds to a singular curve, I guess I expect A to be excellent as well.

flag	asked Oct 22 2009 at 18:29 Benjamin Antieau 2,336●9●18

Bhargav · Accepted Answer · 2009-10-22 18:35:25Z UTC

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Yes: if R is excellent, so is any finite type R-algebra (apply this to Z and A).

answered Oct 22 2009 at 18:35

Bhargav
1,173●5●6

Ah, of course. I was trying to go down from the normalization of A. – Benjamin Antieau Oct 22 2009 at 18:37

Are non-maximal orders in number fields Grothendieck rings?

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