Lemma 2

One has $C \cap m_A \subset pC$.

Choose an integer $m \geq 1$ such that $m_A^m = \{0\}$, whence $C = C_m = C_{m-1}$. Let $\Lambda$ be the subset of $\mathbf{N}^{(S)}$ formed by the families of integers $(\alpha_s)_{s \in S}$ with finite support satisfying $0 \leq \alpha_s < p^m$ for every $s \in S$. Since $B_m$ contains $s^{p^m} = \Phi_m(s, 0, ..., 0)$ for every $s \in S$, the monomials $Z_\alpha = \prod_{s \in S} s^{\alpha_s}$, where $\alpha$ runs through $\Lambda$, generate $C_m$ as a $B_m$-module.

Moreover, from the formula

every element of $B_m$ is of the form $a^{p^m} + pb$ with $a \in A$ and $b \in B_{m-1}$. Consequently every element of $C = C_m$ is of the form with $c_\alpha \in A$ for every $\alpha \in \Lambda$, and $y \in C_{m-1} = C$. If $x$ belongs to $C \cap m_A$, one has $\pi(x) = 0$, whence $\sum_{\alpha \in \Lambda} \pi(c_\alpha)^{p^m} \pi(Z_\alpha) = 0$. Since $\pi(S)$ is a $p$-base of $k$, one has $\pi(c_\alpha) = 0$ for every $\alpha \in \Lambda$ from A, V, p. 96. One then has $c_\alpha \in m_A$, whence $c_\alpha^m = 0$ and a fortiori $c_\alpha^{p^m} = 0$. From (1), one has $x = py$, whence lemma 2.

One has $p^m C = m_A^m = \{0\}$ for $m$ sufficiently large, and assertion b) is therefore trivial. Assertion c) results from lemma 1. If $C'$ is a Cohen subring of $A$ containing $S$, one has $C' \supset C$ from lemma 1. But since the inclusion of $C$ in $C'$ induces an isomorphism of $\kappa_C$ onto $\kappa_{C'}$, one has $C = C'$ (No. 1, prop. 2, b)), and this completes the proving of a).

B) General case

For every integer $n \geqslant 0$, let $A_n$ denote the local ring $A/m_A^{n+1}$, let $m_n = m_A/m_A^{n+1}$ denote its maximal ideal, and let $\pi_n$ denote the canonical homomorphism of $A$ onto $A_n$. By A), there exists a unique Cohen subring $C_n$ of $A_n$ containing $\pi_n(S)$. When $0 \leqslant n \leqslant m$, we denote by $\pi_{n,m}$ the canonical homomorphism of $A_m$ onto $A_n$. By Corollary 2 of Proposition 1 of No. 1, $\pi_{n,m}(C_m)$ is a $p$-ring; one has $\pi_{n,m}(C_m) + m_n = A_n$, hence $\pi_{n,m}(C_m)$ is equal to the Cohen subring $C_n$ of $A_n$. By Proposition 3 of No. 1, the subring $\lim C_n$ of $\lim A_n$ is a $p$-ring. Put $C = \bigcap_{n \in \mathbf{N}} \pi_n^{-1}(C_n)$. Since $C$ is the inverse image of $\lim C_n$ under the isomorphism $a \mapsto (\pi_n(a))_{n \in \mathbf{N}}$ of $A$ onto $\lim A_n$, it is a closed subring of $A$, and a $p$-ring. One has $\pi_n(C) = C_n$ for every $n \in \mathbf{N}$ (No. 1, Proposition 3), and in particular $\pi_0(C) = A_0$, that is to say $\pi(C) = k$. Therefore $C$ is a Cohen subring of $A$.

For every integer $n \geqslant 0$, put $J_n = C \cap m_A^n$. Since the local ring $A$ is separated, one has $\bigcap_{n \in \mathbf{N}} J_n = \{0\}$, and in view of the structure of the ideals of a $p$-ring (No. 1, Proposition 1), every ideal of $C$ of the form $p^kC$ contains one of the $J_n$. Conversely, $J_n$ contains $p^nC$. Consequently, the $pC$-adic topology of $C$ is induced by the $m_A$-adic topology of $A$. This proves b).

Let $A'$ be a closed subring of $A$, containing $S$ and such that $A' + m_A = A$. Since $A'$ is closed, one has $A' = \bigcap_{n \in \mathbf{N}} \pi_n^{-1}(\pi_n(A'))$. One has $\pi_n(A') \supset \pi_n(S)$ and $\pi_n(A') + m_n = A_n$, whence $\pi_n(A') \supset C_n$ by what was seen in A). Finally, one has $\pi_n^{-1}(\pi_n(A')) \supset \pi_n^{-1}(C_n)$, whence $A' \supset C$. This proves c). One deduces the uniqueness of a Cohen subring as in A).