Lemma 2

Let $\mathbf{A}$ be a ring endowed with an endomorphism $\sigma$ satisfying $\sigma(a) \equiv a^p \mod. p \cdot \mathbf{A}$ for every $a \in \mathbf{A}$. Let $n \geqslant 1$ be an integer and let $a_0, ..., a_{n-1}$ be elements of $\mathbf{A}$. Put $u_i = \Phi_i(a_0, ..., a_i)$ for $0 \leqslant i \leqslant n - 1$. Let $u_n$ be an element of $\mathbf{A}$. The following conditions are equivalent:

a) There exists $a_n \in \mathbf{A}$ such that $u_n = \Phi_n(a_0, ..., a_n)$.

b) One has $\sigma(u_{n-1}) \equiv u_n \mod. p^n \cdot \mathbf{A}$.

For $0 \leqslant i \leqslant n - 1$, one has $\sigma(a_i) \equiv a_i^p \mod. p \cdot \mathbf{A}$. According to prop. 1 of no. 1 applied to the case where $J_k = p^k \cdot \mathbf{A}$ (for $k \in \mathbf{N}$) and where $m = 1$, one has the congruence

that is to say

Now, according to formula (2), the relation $u_n = \Phi_n(a_0, ..., a_n)$ is equivalent to

The lemma follows.