Bourbaki

Exercise 40, § 1

Commutative Algebra, IX, p. 54

Let AA be a ring. Let J=P×Q\mathbf{J} = P \times Q be a decomposition of the monoid JJ into a product of submonoids PP and QQ (which are therefore generated by the prime numbers they contain). Suppose that every q∈Qq \in Q is invertible in AA, hence also in U(A)U(A) (p. 52, Exercise 35, e)). a) For every a∈U(A)a \in U(A), the series

εQ(a)=∑q∈Qμ(q)qVqFq(a), \varepsilon_Q(a) = \sum_{q \in Q} \frac{\mu(q)}{q} V_q F_q(a),
where μ\mu denotes the Möbius function (Lie, II, p. 71), is convergent for the product topology of U(A)=AJU(A) = A^J. Thus one defines the additive endomorphism εQ=∑q∈Qμ(q)qVqFq\varepsilon_Q = \sum_{q \in Q} \frac{\mu(q)}{q} V_q F_q of U(A)U(A). For every n∈Jn \in \mathbf{J}, the mapping Φn∘εQ\Phi_n \circ \varepsilon_Q of U(A)U(A) into AA is equal to 0 if n∉Pn \notin P and to Φn\Phi_n if n∈Pn \in P. b) For every q∈Q,q≠1q \in Q, q \neq 1, one has
εQVq=0=FqεQ. \varepsilon_Q V_q = 0 = F_q \varepsilon_Q.
(Use Exercise 36, c), p. 52.) c) Now prove that εQ\varepsilon_Q is an idempotent whose image is the intersection of the kernels of the FqF_q (q∈Q,q≠1q \in Q, q \neq 1). For m∈Jm \in \mathbf{J} and a∈Aa \in A, calculate εQVmτ(a)\varepsilon_Q V_m \tau(a). Deduce that the kernel of εQ\varepsilon_Q is the set of elements of the form ∑n∈QVnτ(an)\sum_{n \in Q} V_n \tau(a_n), with an∈Aa_n \in A. The subset εQU(A)\varepsilon_Q U(A) of U(A)U(A) is stable under addition and multiplication. Endowed with these two operations, it is a commutative ring, with unit εQ(1U(A))\varepsilon_Q(1_{U(A)}), and the mapping εQ:U(A)→εQU(A)\varepsilon_Q : U(A) \to \varepsilon_Q U(A) is a ring homomorphism. d) For every q∈Qq \in Q, let us put eq=1qVqεQFqe_q = \frac{1}{q} V_q \varepsilon_Q F_q. Now prove that one has eq2=eqe_q^2 = e_q and eqeq′=0e_q e_{q'} = 0 if q≠q′q \neq q', and that, for every a∈U(A)a \in U(A) one has
(∗)a=∑q∈Qeq(a), (*) \quad a = \sum_{q \in Q} e_q(a),
where the sum is convergent for the product topology of U(A)U(A). (For () reduce to the case where a=X∈U(R[X])a = X \in U(R[X]), where RR is the subring of QQ consisting of rational numbers with denominator in QQ. It is then enough to apply Φ\Phi and to verify the analogue of () in R[X]JR[X]^J, which results from the formula Φpq∘eq′=Φpqδqq′\Phi_{pq} \circ e_{q'} = \Phi_{pq} \delta_{qq'} if p∈P,q∈Q,q′∈Qp \in P, q \in Q, q' \in Q. The subset eqU(A)e_q U(A) of U(A)U(A) is stable under addition and multiplication. Endowed with these two operations, it is a commutative ring, with unit eq(1U(A))e_q(1_{U(A)}) and the mapping eq:U(A)→eqU(A)e_q : U(A) \to e_q U(A) is a ring homomorphism. e) For every q∈Qq \in Q, one has Fqeq=εQFqF_q e_q = \varepsilon_Q F_q and (1qVq)εQ=eq(1qVq)\left( \frac{1}{q} V_q \right) \varepsilon_Q = e_q \left( \frac{1}{q} V_q \right). Conclude that a↦1qVq(a)a \mapsto \frac{1}{q} V_q(a) is an isomorphism of rings eqU(A)→εQU(A)e_q U(A) \to \varepsilon_Q U(A), with inverse a↦1qVq(a)a \mapsto \frac{1}{q} V_q(a). f) Now prove that the canonical projection πP:U(A)→UP(A)\pi_P : U(A) \to U_P(A) (Exercise 39) defines an isomorphism of rings εQU(A)→UP(A)\varepsilon_Q U(A) \to U_P(A). Deduce an isomorphism of rings of U(A)U(A) onto UP(A)QU_P(A)^Q transforming aa into (πPεQFq(a))q∈Q(\pi_P \varepsilon_Q F_q(a))_{q \in Q} for every a∈U(A)a \in U(A). g) If AA is a Q\mathbf{Q}-algebra, one can take P={1}P = \{1\} and Q=JQ = \mathbf{J}. One deduces an isomorphism of rings U(A)→AJU(A) \to A^J, which is none other than Φ\Phi.

h) Suppose that P={pn∣n∈N}P = \{ p^n | n \in \mathbf{N} \}, where pp is a prime number. Then every prime number q≠pq \neq p is invertible in AA. One obtains an isomorphism of rings U(A)→W(A)QU(A) \to W(A)^Q (cf. exerc. 39, d)). We shall denote by ωA\omega_A the mapping of W(A)W(A) into U(A)U(A) which, by the isomorphism of U(A)U(A) and W(A)QW(A)^Q, corresponds to the inclusion of W(A)W(A) in W(A)QW(A)^Q according to the first component.

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5.4. Not checked by a person.