Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

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

share|improve this answer
    
Ah, of course. I was trying to go down from the normalization of A. –  Benjamin Antieau Oct 22 '09 at 18:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.