Bourbaki

Exercise 41, § 1

Commutative Algebra, IX, p. 55

Let AA be a ring. a) Let μ:A→U(A)\mu : A \to U(A) be a ring homomorphism such that Φ1∘μ=IdA\Phi_1 \circ \mu = \mathrm{Id}_A. Put σ=Φ∘μ\sigma = \Phi \circ \mu and σ(a)=(σn(a))n∈J\sigma(a) = (\sigma_n(a))_{n \in J} for a∈Aa \in A. The mappings σn:A→A\sigma_n : A \to A (n∈Jn \in J) satisfy the following conditions: (1) σn\sigma_n is a ring homomorphism for every n∈Jn \in J, and σ1=IdA\sigma_1 = \mathrm{Id}_A. (2) For every prime number pp and for every a∈Aa \in A, one has σp(a)≡apmod  .pA\sigma_p(a) \equiv a^p \mod. pA and

σp(σn(a))≡σpn(a)mod  .pwp(pn)Afor every n∈J. \sigma_p(\sigma_n(a)) \equiv \sigma_{pn}(a) \mod. p^{w_p(pn)}A \quad \text{for every } n \in J .
b) Suppose that the additive group of AA has no Z\mathbf{Z}-torsion. Let σ=(σn)n∈J\sigma = (\sigma_n)_{n \in J} be a family of mappings σn:A→A\sigma_n : A \to A satisfying conditions (1) and (2) of aa. There exists a unique ring homomorphism μ:A→U(A)\mu : A \to U(A) such that Φ1∘μ=IdA\Phi_1 \circ \mu = \mathrm{Id}_A and Φ∘μ=σ\Phi \circ \mu = \sigma. (Apply Exercises 32 and 33, d), p. 51.) c) Suppose that the additive group of AA has no Z\mathbf{Z}-torsion. There exists a unique ring homomorphism
μA:U(A)→U(U(A)) \mu^A : U(A) \to U(U(A))
such that ΦU(A)∘μA\Phi^{U(A)} \circ \mu^A is the mapping
FA:U(A)→U(A)J F^A : U(A) \to U(A)^J
defined by FA(a)=(Fn(a))n∈JF^A(a) = (F_n(a))_{n \in J}. (Reduce to the case where A=Z[X]A = \mathbf{Z}[X]. Apply bb and Exercises 36 and 38, b), p. 52 and 53.) d) Let X=(Xn)n∈JX = (X_n)_{n \in J} be a family of indeterminates considered as an element of U(Z[X])U(\mathbf{Z}[X]). Put
μZ[X](X)=(μn(X))n∈J=((μn,m(X))m∈J)n∈J \mu^{Z[X]}(X) = (\mu_n(X))_{n \in J} = ((\mu_{n,m}(X))_{m \in J})_{n \in J}
where μn,m(X)∈Z[X]\mu_{n,m}(X) \in \mathbf{Z}[X] for n,mn, m in JJ. For every ring AA define μA:U(A)→U(U(A))\mu^A : U(A) \to U(U(A)) by μA(a)=((μn,m(a))m∈J)n∈J\mu^A(a) = ((\mu_{n,m}(a))_{m \in J})_{n \in J}. Now prove that μA\mu^A is a ring homomorphism such that the following diagram is commutative

U(A)→μAU(U(A))↓ΦA↓ΦU(A)AJ→fA(AJ)J. \begin{array}{ccc} U(A) & \xrightarrow{\mu^A} & U(U(A)) \\ \downarrow \Phi^A & & \downarrow \Phi^{U(A)} \\ A^J & \xrightarrow{f^A} & (A^J)^J . \end{array}

(1)

Here by definition, fA(a)=(fn(a))n∈Jf^A(a) = (f_n(a))_{n \in J} for a∈AJa \in A^J, and ΦA\Phi^A (resp. ΦU(A)\Phi^{U(A)}) is the homomorphism Φ\Phi associated with the ring AA (resp. U(A)U(A)). e) For every ring homomorphism ρ:B→A\rho : B \to A the diagram

U(B)→U(ρ)U(A)↓μB↓μAU(U(B))→U(U(ρ))U(U(A)). \begin{array}{ccc} U(B) & \xrightarrow{U(\rho)} & U(A) \\ \downarrow \mu^B & & \downarrow \mu^A \\ U(U(B)) & \xrightarrow{U(U(\rho))} & U(U(A)) . \end{array}

is commutative.

f) Now prove that the following diagram is commutative

U(A)→μAU(U(A))↓↓μU(A)U(U(A))→U(μA)U(U(U(A))). \begin{array}{ccc} U(A) & \xrightarrow{\mu^A} & U(U(A)) \\ \downarrow & & \downarrow \mu^{U(A)} \\ U(U(A)) & \xrightarrow{U(\mu^A)} & U(U(U(A))) . \end{array}

(Reduce to the case where A=Q[X]A = \mathbf{Q}[X]. Then AA is a Q\mathbf{Q}-algebra, hence ΦA:U(A)→AJ\Phi^A : U(A) \to A^J is an isomorphism and U(A)U(A) is also a Q\mathbf{Q}-algebra. If, with the aid of diagram (1) and by transport of structure, one replaces μA\mu^A by fAf^A when AA is a Q\mathbf{Q}-algebra, one is reduced to proving the commutativity of the diagram

AJ→fA(AJ)J↓fA↓f(AJ)(AJ)J→(fA)J((AJ)J)J. \begin{array}{ccc} A^J & \xrightarrow{f^A} & (A^J)^J \\ \downarrow f^A & & \downarrow f^{(A^J)} \\ (A^J)^J & \xrightarrow{(f^A)^J} & ((A^J)^J)^J . \end{array}

One has

f(AJ)(fA(X))=f(AJ)((fn(X))n∈J)=((fm(fn(X)))n∈J)m∈J=((fmn(X))n∈J)m∈J, f^{(A^J)}(f^A(X)) = f^{(A^J)}((f_n(X))_{n \in J}) = ((f_m(f_n(X)))_{n \in J})_{m \in J} = ((f_{mn}(X))_{n \in J})_{m \in J} ,
and
(fA)J(fA(X))=(fA)J((fn(X))n∈J)=(fA(fn(X)))n∈J=((fmn(X))m∈J)n∈J. (f^A)^J(f^A(X)) = (f^A)^J((f_n(X))_{n \in J}) = (f^A(f_n(X)))_{n \in J} = ((f_{mn}(X))_{m \in J})_{n \in J} .

g) For all a∈Aa \in A, one has

μnA(τ(a))=0forn≥2. \mu_n^A(\tau(a)) = 0 \quad \text{for} \quad n \geq 2 .

h) One has the following commutative diagram

U(A)→μAU(U(A))↓φ↓ψW(A)→sAW(W(A)) \begin{array}{ccc} U(A) & \xrightarrow{\mu^A} & U(U(A)) \\ \downarrow \varphi & & \downarrow \psi \\ W(A) & \xrightarrow{s_A} & W(W(A)) \end{array}

where the mapping sAs_A is that defined in Exercise 15, p. 44, where the mapping φ\varphi from U(A)U(A) into W(A)W(A) is obtained by identifying W(A)W(A) and Up∞(A)U_{p^\infty}(A) (p. 53, Exercise 39), and the mapping ψ\psi is the composite of U(φ)U(\varphi) and the mapping from U(W(A))U(W(A)) into W(W(A))W(W(A)) obtained by identifying W(W(A))W(W(A)) with Up∞(W(A))U_{p^\infty}(W(A)).)

Solution

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