In the excellent book "Algebraic Geometry 1" of Görtz & Wedhorn, in exercise 3.14, one is asked to show that in the spectrum of a valuation ring with infinite infinitely many primes, the complement of the maximal ideal unique closed point is an open set without a closed point. It seems to me this is quite not true : to find a counter example, it is sufficient to build a valuation ring with a numerable descending chain of prime ideals, which can be done by using as value group $\mathbb{Z}^{\mathbb{N}}$ with the reverse lexicographical order. Did I miss something ?
|
2 | corrected grammar, clarified meaning according to the OP's comment | ||
|
|
||||
|
|
1 |
|
||
Closed points of valuation schemeIn the excellent book "Algebraic Geometry 1" of Görtz & Wedhorn, in exercise 3.14, one is asked to show that in the spectrum of a valuation ring with infinite many primes, the complement of the maximal ideal is an open set without a closed point. It seems to me this is quite not true : to find a counter example, it is sufficient to build a valuation ring with a numerable descending chain of prime ideals, which can be done by using as value group $\mathbb{Z}^{\mathbb{N}}$ with the reverse lexicographical order. Did I miss something ?
|
||||
