Bourbaki

Exercise 8, § 4

Commutative Algebra, III, 258

Let AA be a linearly topologized commutative ring. The additive group A[[X1,…,Xp]]A[[X_1, \ldots, X_p]] is identified with the product group ANpA^{N^p} and is given the product topology T\mathscr{T}.

(a) Show that T\mathscr{T} is compatible with the ring structure on A[[X1,…,Xp]]A[[X_1, \ldots, X_p]] and is a linear topology for which the mappings f↦∂f/∂Xif \mapsto \partial f / \partial X_i are continuous.

(b) For every element P=∑eαe,PXeP = \sum_e \alpha_{e,P} X^e of A[[X1,…,Xp]]A[[X_1, \ldots, X_p]] (notation of § 1, no. 6) and every complete Hausdorff topological A-algebra B (§2§ 2, Exercise 28), an element x=(x1,…,xr)∈Bpx = (x_1, \ldots, x_r) \in B^p is said to be substitutable in PP if the family (αe,Pxe)(\alpha_{e,P} x^e) (where we have written xe=x1e1…xpepx^e = x_1^{e_1} \ldots x_p^{e_p} for e=(e1,…,ep)e = (e_1, \ldots, e_p)) converges to 0 in B with respect to the filter of complements of finite subsets of NpN^p. This family is then summable and its sum is denoted by P(x)P(x). Show that the set of formal power series P∈A[[X1,…,Xp]]P \in A[[X_1, \ldots, X_p]] such that xx is substitutable in PP is a subring S, of A[[X1,…,Xp]]A[[X_1, \ldots, X_p]] and the mapping P↦P(x)P \mapsto P(x) is a homomorphism from S, to B.

(c) Show that, if xx is substitutable in PP and y=(y1,…,yp)y = (y_1, \ldots, y_p) is an element of BpB^p such that the yiy_i are topologically nilpotent for 1≤i≤p1 \leq i \leq p, then x+yx + y is substitutable in PP.

In particular, if there exists in B a neighbourhood of 0 consisting of topologically nilpotent elements, then, for all P∈A[[X1,…,Xp]]P \in A[[X_1, \ldots, X_p]], the set D of x∈Bpx \in B^p which are substitutable in PP is open and the mapping x↦P(x)x \mapsto P(x) from D to B is continuous.

(d) Suppose from now on that A is Hausdorff and complete and let A[[X1,…,Xp]]A[[X_1, \ldots, X_p]] be given the topology T\mathscr{T}. For a system of q formal power series P1,…,PqP_1, \ldots, P_q of A[[X1,…,Xp]]A[[X_1, \ldots, X_p]] to be substitutable in a formal power series

Q∈A[[X1,…,Xq]], Q \in A[[X_1, \ldots, X_q]],

it is necessary and sufficient that the system (P1(0),…,Pq(0))(P_1(0), \ldots, P_q(0)) be substitutable in Q.

(e) Suppose that xx is substitutable in each of the PkP_k (1≤k≤q1 \leq k \leq q) and that the system (P1,…,Pq)(P_1, \ldots, P_q) is substitutable in Q. Show that xx is substitutable in Q(P1,…,Pq)Q(P_1, \ldots, P_q), that the system (P1(x),…,Pq(x))(P_1(x), \ldots, P_q(x)) is substitutable in Q and that Q(P1(x),…,Pq(x))=((Q(P1,…,Pq))(x))Q(P_1(x), \ldots, P_q(x)) = ((Q(P_1, \ldots, P_q))(x)) iff further one of the following hypotheses assumed to be satisfied: (a) there is in B an ideal n such that the ordered pair (B,n)(B, n) satisfies Hensel's conditions and the xix_i (1≤i≤p1 \leq i \leq p) belong to n; (β) the formal power series Q is restricted.

(f) Take B=A=F2B = A = \mathbf{F}_2 with the discrete topology, P=X+X2P = X + X^2, Q=∑n=0∞X2nQ = \sum_{n=0}^\infty X^{2^n}; then PP is substitutable in QQ, x=1x = 1 is substitutable in PP and in Q(P)Q(P) and P(x)P(x) is substitutable in QQ, but Q(P(x))=0Q(P(x)) = 0 and (Q(P))(x)=1(Q(P))(x) = 1.

Solution

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