Bourbaki

Exercise 15, § 1

Commutative Algebra, IX, p. 44

a) Let A be a ring such that multiplication by p⋅1Ap \cdot 1_A in A is injective. Then there exists a unique ring homomorphism sAs_A of W(A)W(A) into W(W(A))W(W(A)) satisfying

sA∘FA=FW(A)∘sA s_A \circ F_A = F_{W(A)} \circ s_A
and
Φ0∘sA=IdW(A) \Phi_0 \circ s_A = \mathrm{Id}_{W(A)}
(where Φ0\Phi_0 is the projection of W(W(A))W(W(A)) onto W(A)W(A)). It is also the unique ring homomorphism satisfying Φn∘sA=FAn\Phi_n \circ s_A = F_A^n for every integer n∈Nn \in \mathbf{N} (where Φn:W(W(A))→W(A)\Phi_n : W(W(A)) \to W(A) is the nn-th ghost component in W(W(A))W(W(A))). b) Consider the ring A=Z[(Xn)n∈N]\mathscr{A} = \mathbf{Z}[(X_n)_{n \in \mathbf{N}}] of polynomials in a family of indeterminates (Xn)n∈N(X_n)_{n \in \mathbf{N}}. Let X be the element (Xn)n∈N(X_n)_{n \in \mathbf{N}} of W(A)W(\mathscr{A}). Put sA(X)=(sn(X))n∈Ns_{\mathscr{A}}(X) = (s_n(X))_{n \in \mathbf{N}}, where sn(X)∈W(A)s_n(X) \in W(\mathscr{A}). For

EXERCISES

every ring A, define the mapping sAs_A of W(A)W(A) into W(W(A))W(W(A)) by the formula sA(a)=(sn(a))n∈Ns_A(a) = (s_n(a))_{n \in \mathbf{N}}. Prove that sAs_A is a ring homomorphism satisfying

sA∘FA=FW(A)∘sA s_A \circ F_A = F_{W(A)} \circ s_A

and

Φ0∘sA=IdW(A). \Phi_0 \circ s_A = \mathrm{Id}_{W(A)}.

c) For every ring homomorphism u:B→Au : B \to A, one has

sA∘W(u)=W(W(u))∘sB. s_A \circ W(u) = W(W(u)) \circ s_B.

d) The mappings W(sA)∘sAW(s_A) \circ s_A and sW(A)∘sAs_{W(A)} \circ s_A of W(A)W(A) into W(W(W(A)))W(W(W(A))) are equal.

e) For every x∈Ax \in A, one has sA(τA(x))=τW(A)(τA(x))s_A(\tau_A(x)) = \tau_{W(A)}(\tau_A(x)).

f) One has sA∘VA=VW(A)∘sAs_A \circ V_A = V_{W(A)} \circ s_A, and the mapping sAs_A is continuous when W(A)W(A) and W(W(A))W(W(A)) are endowed with the G\mathscr{G} topologies.

Solution

No solution has been written yet.

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