Exercise 23, § 3
Commutative Algebra, VII, 569
Bourbaki marks this one as harder than the rest.
Let be a Krull domain, its field of fractions, a derivation of such that , the subfield of the kernel of and the Krull domain ; suppose that is of characteristic ; then and , so that is the integral closure of in and the canonical homomorphism is defined (, no. 10). Let denote the group of invertible elements of .
(a) If is such that is the canonical image of a divisor in , show that (note that for every prime ideal of of height 1 there exists such that and observe that is invariant under ). Let be the additive subgroup of consisting of the ("logarithmic derivatives") which belong to (for or , which amounts to the same, and ) and let be the subgroup of consisting of the , where . Show that, for every divisor such that the image of is a principal divisor in , the class mod. of for all such that depends only on the class of in and derive a canonical injective homomorphism of to .
(b) Show that, if is contained in no prime ideal of of height 1 and , is bijective. (Reduce it to showing that, if is such that and is a prime ideal of of height 1 such that is not a multiple of , then , where ; for this, deduce from the hypothesis that, if is a uniformizer of , then , so that is invariant under and that therefore defines by taking quotients a derivation of the residue field ; show that and deduce that )
(c) Let be a field of characteristic 2, the polynomial ring and the derivation of such that ; the field the kernel of is such that . Then , but show that is not the canonical image of a divisor in . (Argue by reductio ad absurdum supposing that for ; there would then be in a uniformizer of , necessarily of the form where ; deduce that and obtain a contradiction by calculating .)
Solution
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