Bourbaki

Exercise 9, § 1

Commutative Algebra, VII, 547

(a) Let BB be a discrete valuation ring, A=B[[X]]A = B[[X]] the ring of formal power series in one indeterminate over BB and C=AC = A, the ring of fractions S−1AS^{-1}A, where SS is the multiplicative set of Xh (h≥0)X^h \ (h \geq 0). Let vv be a normed valuation on BB; for every element f=∑n≥hbnXn≠0f = \sum_{n \geq h} b_n X^n \neq 0 of CC with bh≠0b_h \neq 0 in BB (hh positive or negative), we write s(f)=v(bh)s(f) = v(b_h). Show that ss is a Euclidean stathm on CC (Algebra, Chapter VII, § 1, Exercise 7), such that s(fg)=s(f)+s(g)s(fg) = s(f) + s(g) for non-zero f,gf, g in CC. (If s(f)=p≥s(g)=qs(f) = p \geq s(g) = q and hh is the least of the degrees of the terms ≠0\neq 0 in ff, show that there exists u∈Cu \in C such that f−ug=Xh+1f1f - ug = X^{h+1}f_1 where f1∈Cf_1 \in C and, by arguing by reductio ad absurdum deduce the existence of a process of "Euclidean division" in CC.) If s(f)=0s(f) = 0, ff is invertible in CC.

Show that AA is not a Dedekind domain.

(b) Deduce from (a) that, if BB is a Krull domain, A=B[[X]]A = B[[X]] is also a Krull domain. (If KK is the field of fractions of BB, note that AA is the intersection of K[[X]]K[[X]] and a family (C_i) of principal ideal domains with the same field of fractions as AA and that every element of AA is invertible in almost all the CiC_i.)

Solution

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