Proposition 6

Let $A$ be a ring.

a) For every integer $k \geqslant 1$, one has $(V_1(A))^k = p^{k-1} \cdot V_1(A)$.

b) Suppose that $A$ is a ring of characteristic $p$. On the ring $W(A)$, the $V_1(A)$-adic topology and the $p$-adic topology coincide, and they are finer than the product topology $\mathscr{C}$ (cf. No. 6). The ring $W(A)$ is separated and complete for the $p$-adic topology.

Let us prove a) by induction on $k$. The case $k = 1$ is evident. Suppose $k \geqslant 2$. By the induction hypothesis, one has $V_1(A)^{k-1} = p^{k-2} \cdot V_1(A)$ and consequently $V_1(A)^k = p^{k-2} \cdot (V_1(A))^2$. But it follows from prop. 3, d), formula (31), of No. 5 that one has $(V_1(A))^2 = p \cdot V_1(A)$, whence a).

Suppose now that $A$ is of characteristic $p$. As one has

one deduces from a) the inclusions $p^k.W(A) \subset (V_1(A))^k \subset p^{k-1}.W(A)$, and from the corollary to prop. 5 the inclusion $(V_1(A))^k \subset V_k(A)$, for every integer $k \geqslant 1$. The first assertion of b) follows.

Let $k$ be an integer $\geqslant 1$. According to formula (52), the ideal $p^k.W(A)$ of $W(A)$ is the set of elements $a = (a_n)_{n \in \mathbf{N}}$ of $W(A)$ such that one has $a_n = 0$ for $n < k$ and $a_n \in A^{p^k}$ for $n \geqslant k$. It is therefore closed for the topology $\mathscr{G}$. Since $W(A)$ is separated and complete for the topology $\mathscr{G}$ (no 6) and since the ideals $p^k.W(A)$ of $W(A)$, for $k \geqslant 1$, form a basis of neighbourhoods of $0$ in $W(A)$ for the $p$-adic topology, the ring $W(A)$ is separated and complete for the $p$-adic topology (TG, III, p. 26, cor. 1 to prop. 10).