Bourbaki

Exercise 23, § 3

Commutative Algebra, VII, 569

Bourbaki marks this one as harder than the rest.

Let BB be a Krull domain, EE its field of fractions, AA a derivation of EE such that Δ(B)⊂B\Delta(B) \subset B, KK the subfield of EE the kernel of AA and AA the Krull domain B∩KB \cap K; suppose that KK is of characteristic p>0p > 0; then Ep⊂KE^p \subset K and Bp⊂AB^p \subset A, so that BB is the integral closure of AA in EE and the canonical homomorphism i~:C(A)→C(B)\tilde{i}: C(A) \to C(B) is defined (§1§ 1, no. 10). Let UU denote the group of invertible elements of BB.

(a) If b∈Eb \in E is such that div⁡B(b)\operatorname{div}_B(b) is the canonical image of a divisor in D(A)D(A), show that Δb/b∈B\Delta b/b \in B (note that for every prime ideal P\mathfrak{P} of BB of height 1 there exists b′∈Kb' \in K such that vP(b)=vP(b′)v_{\mathfrak{P}}(b) = v_{\mathfrak{P}}(b') and observe that B1B_1 is invariant under AA). Let LL be the additive subgroup of BB consisting of the Δb/b\Delta b/b ("logarithmic derivatives") which belong to BB (for b∈Eb \in E or b∈Bb \in B, which amounts to the same, and b≠0b \neq 0) and let L′L' be the subgroup of LL consisting of the Δu/u\Delta u/u, where u∈Uu \in U. Show that, for every divisor d∈D(A)d \in D(A) such that the image of dd is a principal divisor in D(B)D(B), the class mod. L′L' of Δb/b\Delta b/b for all bb such that i(d)=div⁡B(b)i(d) = \operatorname{div}_B(b) depends only on the class of dd in C(A)C(A) and derive a canonical injective homomorphism †\dagger of Ker⁡(i~)\operatorname{Ker}(\tilde{i}) to L/L′L/L'.

(b) Show that, if Δ(B)\Delta(B) is contained in no prime ideal of BB of height 1 and [E:K]=p[E: K] = p, ϕ\phi is bijective. (Reduce it to showing that, if b∈Eb \in E is such that Δb/b∈B\Delta b/b \in B and P\mathfrak{P} is a prime ideal of BB of height 1 such that vP(b)v_{\mathfrak{P}}(b) is not a multiple of pp, then e(P/p)=1e(\mathfrak{P}/p) = 1, where p=P∩Ap = \mathfrak{P} \cap A; for this, deduce from the hypothesis that, if tt is a uniformizer of BPB_{\mathfrak{P}}, then Δt/t∈BP\Delta t/t \in B_{\mathfrak{P}}, so that PBP\mathfrak{P}B_{\mathfrak{P}} is invariant under AA and that AA therefore defines by taking quotients a derivation Δ‾\overline{\Delta} of the residue field k=BP/PBPk = B_{\mathfrak{P}}/\mathfrak{P}B_{\mathfrak{P}}; show that Δ‾≠0\overline{\Delta} \neq 0 and deduce that f(P/p)=p.f(\mathfrak{P}/p) = p.)

(c) Let kk be a field of characteristic 2, BB the polynomial ring k[X,Y,Z]k[X, Y, Z] and AA the derivation of E=k(X,Y,Z)E = k(X, Y, Z) such that Δ(X)=Y4,Δ(Y)=X2,Δ(Z)=XYZ\Delta(X) = Y^4, \Delta(Y) = X^2, \Delta(Z) = XYZ; the field KK the kernel of AA is such that [E:K]=4[E : K] = 4. Then Δ(Z)/Z∈B\Delta(Z)/Z \in B, but show that div⁡B(Z)\operatorname{div}_B(Z) is not the canonical image of a divisor in D(A)D(A). (Argue by reductio ad absurdum supposing that e(P/p)=1e(\mathfrak{P}/p) = 1 for P=BZ,p=P∩A\mathfrak{P} = BZ, p = \mathfrak{P} \cap A; there would then be in AA a uniformizer of BB, necessarily of the form a(X,Y,Z)Za(X, Y, Z)Z where b=a(X,Y,0)≠0b = a(X, Y, 0) \neq 0; deduce that Δb/b=−XY\Delta b/b = -XY and obtain a contradiction by calculating A(−XY)A(-XY).)

Solution

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Transcribed from the printed volume. Not translated.