Lemma 1

Let $A'$ be a subring of $A$ containing $S$. In order that $A'$ contain $C_n$, it is necessary and sufficient that one have $A' + m_A = A$.

One has $pA \subset m_A$ and $B_n$ consists of the elements of the form $a_0^{p^n} + pa_1^{p^{n-1}} + \cdots + p^n a_n$ with $a_0, ..., a_n$ in $A$. Hence one has $\pi(B_n) = k^{p^n}$, whence $\pi(C_n) = k^{p^n}[\pi(S)]$. But since $\pi(S)$ is a $p$-basis of $k$, one has $k = k^{p^n}[\pi(S)]$ (A, V, p. 96), whence $\pi(C_n) = k$, that is to say $C_n + m_A = A$.

Let $A'$ be a subring of $A$ containing $S$. If $A'$ contains $C_n$, we have

Conversely, suppose that one has $A' + m_A = A$. Let $a_0, ..., a_n$ be elements of $A$; by assumption there exist elements $a'_0, ..., a'_n$ of $A'$ such that $a_i \equiv a'_i \mod m_A$ for $0 \leq i \leq n$. By prop. 1 of § 1, No. 1 and the assumption $m_A^{n+1} = \{0\}$, we therefore have $\Phi_n(a_0, ..., a_n) = \Phi_n(a'_0, ..., a'_n) \in A'$, whence $B_n \subset A'$. Since $C_n$ is the ring generated by $B_n \cup S$, one has $C_n \subset A'$.

In the set $S$ of subrings $A'$ of $A$ containing $S$ and such that $A' + m_A = A$, there exists by lemma 1 a smallest element $C$, and one has $C_n = C$ for every integer $n \geq 0$ such that $m_A^{n+1} = \{0\}$.

One has $C + m_A = A$ by construction and $p1_C$ is nilpotent. One obviously has $pC \subset C \cap m_A$ and lemma 2 below therefore shows that $pC$ is a maximal ideal of $C$ and consequently that $C$ is a Cohen subring of $A$.