Bourbaki

Proposition 4

Tag 0CUC, permanent. Cite it as [Bourbaki, Tag 0CUC].

Commutative Algebra, chapter IX, § 1. Vecteurs de Witt

Let aa and bb be in AA and let x=(xn)n∈Nx = (x_n)_{n \in \mathbf{N}} be an element of W(A)W(A).

a) We have the formulas

τ(ab)=τ(a)×τ(b) \tau(ab) = \tau(a) \times \tau(b)
τ(a)×x=(apnxn)n∈N. \tau(a) \times x = (a^{p^n} x_n)_{n \in \mathbf{N}}.

b) The series with general term Vn(τ(xn))V^n(\tau(x_n)) is convergent in W(A)W(A), with sum xx. Let nn be a positive integer. The polynomial Pn(X0,...,Xn;Y0,...,Yn)P_n(X_0, ..., X_n; Y_0, ..., Y_n) introduced in No. 3 is isobaric of weight pnp^n in the family (X0,...,Xn)(X_0, ..., X_n) when XiX_i is assigned the weight pip^i. Hence

Pn(X0,0,...,0;Y0,...,Yn)=X0pnPn(1,0,...,0;Y0,...,Yn). P_n(X_0, 0, ..., 0; Y_0, ..., Y_n) = X_0^{p^n} P_n(1, 0, ..., 0; Y_0, ..., Y_n).
Since 1=(1,0,0,...)1 = (1, 0, 0, ...) is the unit element of the ring of Witt vectors with coefficients in Z[(Xn)n∈N,(Yn)n∈N]\mathbf{Z}[(X_n)_{n \in \mathbf{N}}, (Y_n)_{n \in \mathbf{N}}], we have
Pn(1,0,...,0;Y0,...,Yn)=Yn. P_n(1, 0, ..., 0; Y_0, ..., Y_n) = Y_n.
By substituting aa for X0X_0 and xix_i for YiY_i, we deduce from (40) and (41) the relation:
Pn(a,0,...,0;x0,...,xn)=apnxn. P_n(a, 0, ..., 0; x_0, ..., x_n) = a^{p^n} x_n.
According to the definition of the multiplication in W(A)W(A), we have proved (39); formula (38) is a particular case of (39).

Let us prove b). By definition, Vn(τ(xn))V^n(\tau(x_n)) is the sequence all of whose components are zero, except the one with index nn, which is equal to xnx_n. It follows from Lemma 4, by induction on mm, that we have

∑n=0mVn(τ(xn))=(x0,...,xm,0,0,...) \sum_{n=0}^m V^n(\tau(x_n)) = (x_0, ..., x_m, 0, 0, ...)

for every integer m≥0m \geq 0; we deduce b) by passing to the limit since the topology T\mathscr{T} on W(A)W(A) is the product of the discrete topologies of the factors AA.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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