Bourbaki

Exercise 37, § 1

Commutative Algebra, IX

Let A be a ring. a) Let m∈Jm \in \mathbf{J}. For every element a=(an)n∈Ja = (a_n)_{n \in \mathbf{J}} of U(A), one has

a=(a1,...,am,0,...,0,...)+(0,...,0⏟m terms,am+1,am+2,...). a = (a_1, ..., a_m, 0, ..., 0, ...) + (\underbrace{0, ..., 0}_{m \text{ terms}}, a_{m+1}, a_{m+2}, ...).
b) We endow U(A) with the product topology on AJA^\mathbf{J} of the discrete topology on each of the factors. It makes U(A) into a separated and complete topological ring. c) We denote by τA\tau_A (or τ\tau) the mapping from A into U(A) which associates to a∈Aa \in A the element (a,0,...,0)∈U(A)(a, 0, ..., 0) \in U(A). Let a,ba, b be two elements of A, x=(xn)n∈Jx = (x_n)_{n \in \mathbf{J}} an element of U(A). (i) One has the formulae
τ(ab)=τ(a)×τ(b),τ(a)×x=(anxn)n∈J, \tau(ab) = \tau(a) \times \tau(b), \quad \tau(a) \times x = (a^n x_n)_{n \in \mathbf{J}},
Φ(τ(a))=(an)n∈J, \Phi(\tau(a)) = (a^n)_{n \in \mathbf{J}},
Fn(τ(a))=τ(an)for every n∈J. F_n(\tau(a)) = \tau(a^n) \quad \text{for every } n \in \mathbf{J}.
(ii) The series with general term Vnτ(xn)V_n \tau(x_n) is convergent in U(A), with sum x.

Solution

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