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 infinitely many primes, the complement of the 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 ?
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3$\begingroup$ Can you explain your example a little more? Namely, why do you think this is a counterexample? If it is a valuation ring, then it is local so there is a unique maximal ideal. This is "the" closed point. Your other prime ideals are "points" but they are not closed. Mostly I guess I'm asking what other closed point is in your ring? $\endgroup$– MattCommented May 22, 2011 at 0:13
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1$\begingroup$ @Matt The point of the exercise as I explained is to look at closed points in the open set which is precisely the complement of the maximal ideal ! If there is one, it will not be closed obviously in the whole spectrum but his closure will contain itself and the maximal ideal. So in my counter example, the complement of the maximal ideal is an open set which contains one closed point only, which is the largest prime contained in the maximal one. $\endgroup$– brunohCommented May 22, 2011 at 7:25
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7$\begingroup$ Your counterexample is correct. The statement in the book you mention is only true, if the maximal ideal of the valuation ring is the union of all prime ideals properly contained in it - such a prime ideal is called a limit prime. H $\endgroup$– HagenCommented May 23, 2011 at 15:11
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3$\begingroup$ @Hagen You have given a perfectly clear and right answer to my question. Thank you very much for taking the time. How can I award you reputation points or mark my question as answered ? Sorry but I am a bit new to this ... $\endgroup$– brunohCommented May 23, 2011 at 15:38
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8$\begingroup$ Thank you very much for the pointing out the mistake in the exercise. I will post it in the list of errata on www.algebraic-geometry.de $\endgroup$– Torsten WedhornCommented May 27, 2011 at 7:48
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The answer to my question has been given by the comment of Hagen Knaf