Example 3

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

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

Let $n \geq 1$ be an integer. If $a_0, ..., a_{n-1}, b_0, ..., b_{n-1}$ are integers such that $a_i \equiv b_i \bmod p$ for $0 \leq i < n$ , one has (No. 1, prop. 1)

\Phi_{n-1}(a_0, ..., a_{n-1}) \equiv \Phi_{n-1}(b_0, ..., b_{n-1}) \bmod p^n.

Consequently, $\Phi_{n-1}$ defines by passing to the quotients a ring homomorphism $\varphi_n : W_n(\mathbf{Z}/p\mathbf{Z}) \to \mathbf{Z}/p^n\mathbf{Z}$ . The image of $\varphi_n$ is a subgroup of $\mathbf{Z}/p^n\mathbf{Z}$ containing 1, hence $\varphi_n$ is surjective. Since the finite sets $W_n(\mathbf{Z}/p\mathbf{Z})$ and $\mathbf{Z}/p^n\mathbf{Z}$ have the same cardinal $p^n$ , $\varphi_n$ is an isomorphism.

Let $m$ and $n$ be integers such that $1 \leq n \leq m$ . There exists a unique ring homomorphism $\alpha_{n,m} : \mathbf{Z}/p^m\mathbf{Z} \to \mathbf{Z}/p^n\mathbf{Z}$ ; consequently the diagram

\begin{array}{ccc} \mathbf{Z}/p^m\mathbf{Z} & \xrightarrow{\alpha_{n,m}} & \mathbf{Z}/p^n\mathbf{Z} \\ \varphi_m \uparrow & & \varphi_n \uparrow \\ W_m(\mathbf{Z}/p\mathbf{Z}) & \xrightarrow{\pi_{n,m}} & W_n(\mathbf{Z}/p\mathbf{Z}) \end{array}

is commutative. It follows that $\varphi = \lim \varphi_n$ is an isomorphism of topological rings from $W(\mathbf{Z}/p\mathbf{Z}) = \lim_{\leftarrow} W_n(\mathbf{Z}/p\mathbf{Z})$ onto $\mathbf{Z}_p = \lim_{\leftarrow} \mathbf{Z}/p^n\mathbf{Z}$ (III, § 2, No. 12, example 3).

Let $m$ and $n$ be two integers $\geq 1$ . By construction, one has an exact sequence of additive groups

(E)

0 \longrightarrow W(A) \xrightarrow{\nu^m} W(A) \xrightarrow{\pi_m} W_m(A) \longrightarrow 0.

By passing to the quotients, the endomorphism $V^n$ of the additive group of $W(A)$ defines a homomorphism $V^n_m$ of the additive group of $W_m(A)$ into that of $W_{m+n}(A)$ . In other words, one has a commutative diagram

\begin{array}{ccc} W(A) & \xrightarrow{V^n} & W(A) \\ \pi_m \downarrow & & \pi_{n+m} \downarrow \\ W_m(A) & \xrightarrow{V^n_m} & W_{n+m}(A) \end{array}

By passing to the quotients, one deduces from the exact sequence (E) an exact sequence

(E')

0 \longrightarrow W_m(A) \xrightarrow{V^n_m} W_{n+m}(A) \xrightarrow{\pi_{n,n+m}} W_n(A) \longrightarrow 0 .

One has

(45)

V^n_m[a_0, ..., a_{m-1}] = [\underbrace{0, ..., 0}_{n \text{ fois}}, a_0, ..., a_{m-1}],

for every element $[a_0, ..., a_{m-1}]$ of $W_m(A)$ .

According to prop. 3, c) of No. 5, one has $FV^{m+1}(a) = p.V^m(a)$ for every $a$ in $W(A)$ and one consequently has $F(V_{m+1}(A)) \subset V_m(A)$ . By induction on $n$ , one deduces that $F^n$ maps $V_{n+m}(A)$ into $V_m(A)$ , and hence defines, by passing to the quotients, a ring homomorphism $F^n_m : W_{n+m}(A) \to W_m(A)$ . By construction, one has a commutative diagram

\begin{array}{ccc} W(A) & \xrightarrow{F^n} & W(A) \\ \pi_{n+m} \downarrow & & \pi_m \downarrow \\ W_{n+m}(A) & \xrightarrow{F^n_m} & W_m(A) \end{array}

Recall (No. 3) that the polynomial $F_i$ belongs to $\mathbf{Z}[X_0, ..., X_{i+1}]$ for every integer $i \geqslant 0$ ; the homomorphism $F^1_m$ from $W_{m+1}(A)$ into $W_m(A)$ is therefore made explicit as follows:

(46)

F^1_m[a_0, ..., a_m] = [F_i(a_0, ..., a_{i+1})]_{0 \leqslant i < m}.

Let $a \in W_m(A)$ , $a' \in W_m(A)$ and $b \in W_{m+1}(A)$ . The following formulas result by passing to the quotients from prop. 3 of No. 5:

(47)

F^1_m(V^1_m(a)) = p.a

(48)

V^1_m(a \times F^1_m(b)) = V^1_m(a) \times b

(49)

V^1_m(a) \times V^1_m(a') = p.V^1_m(a \times a')

(50)

V^1_m(F^1_m(b)) = \mu_{m+1} \times b

(avec $\mu_{m+1} = [0, 1, 0, ..., 0]$ ).

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

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