Bourbaki

Exercise 1, § 4

Commutative Algebra, VI, 449

(a) Let G be a totally ordered group and M a major set in G, not containing 0. Show that there exists a greatest isolated subgroup H of G not meeting M.

(b) Let A be a valuation ring and vv a valuation associated with A. For an ideal ϱ≠0\varrho \neq 0 of A to be prime, it is necessary and sufficient that p correspond to a major set M in the totally ordered value group G of vv such that, if H is the greatest isolated subgroup of G not meeting M, M is the complement of H+H_+ in G,.

(c) With the notation of (b), for an ideal q of A to be p-primary, it is necessary and sufficient that it satisfy one of the following conditions: either H={0}H = \{0\} (and p is then maximal) or q=ϱq = \varrho.

(d) For an ideal a of A which is distinct from 0 and A to satisfy a2=aa^2 = a, it is necessary and sufficient that a correspond to a major set M such that, if H is the greatest isolated subgroup of G not meeting M, the image M′M' of M in the totally ordered group G′=G/HG' = G/H has no least element; a is then a prime ideal.

Solution

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Transcribed from the printed volume. Not translated.