Bourbaki

Exercise 8, § 4

Commutative Algebra, X

Bourbaki marks this one as harder than the rest.

Let AA be a regular local ring, x=(x1,…,xd)x = (x_1, \ldots, x_d) a system of coordinates of AA, ρ:A→B\rho : A \to B an injective ring homomorphism making BB a finitely generated AA-module. Prove that the following conditions are equivalent:

(i) for every integer pp, the element ρ(x1…xd)p\rho(x_1 \ldots x_d)^p does not belong to the ideal of BB generated by ρ(x1p+1),…,ρ(xdp+1)\rho(x_1^{p+1}), \ldots, \rho(x_d^{p+1}); (ii) the AA-module ρ(A)\rho(A) is a direct factor of BB.

These conditions are satisfied if AA is a Q\mathbf{Q}-algebra (exercise 3 of § 2).

(To prove (ii) ⇒\Rightarrow (i), adapt the proof of loc. cit. For the opposite implication, reduce to the case where AA is complete, and set for p⩾0p \geqslant 0 Ap=A/(x1p+1,…,xdp+1)A_p = A/(x_1^{p+1}, \ldots, x_d^{p+1}) and Bp=B/(x1p+1,…,xdp+1)BB_p = B/(x_1^{p+1}, \ldots, x_d^{p+1})B. Observe that the class of (x1…xd)p(x_1 \ldots x_d)^p generates the socle of ApA_p, and deduce that the homomorphism ρp:Ap→Bp\rho_p : A_p \to B_p induced by ρ\rho is injective. Prove that ρp\rho_p admits an ApA_p-linear retraction. Conclude with the aid of E, III, p. 60, example II.)

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.