Proposition 2

Let $\mathbf{A}$ be a ring.

(a) If $p \cdot 1_A$ is not a divisor of 0 in $\mathbf{A}$, the mapping $\Phi_A$ is injective.

(b) If $p \cdot 1_A$ is invertible in $\mathbf{A}$, the mapping $\Phi_A$ is bijective.

(c) If $\sigma$ is an endomorphism of the ring $\mathbf{A}$, satisfying $\sigma(a) \equiv a^p \mod. p \cdot \mathbf{A}$ for every $a \in \mathbf{A}$, the image $\mathbf{A}'$ of $\Phi_A$ is a subring of $\mathbf{A}^{\mathbf{N}}$, stable under $f_A$ and $v_A$. It is the set of elements $(u_n)_{n \in \mathbf{N}}$ of $\mathbf{A}^{\mathbf{N}}$ such that $\sigma(u_n) \equiv u_{n+1} \mod. p^{n+1} \cdot \mathbf{A}$ for every $n \in \mathbf{N}$.

If $a = (a_n)_{n \in \mathbf{N}}$ and $u = (u_n)_{n \in \mathbf{N}}$ are elements of $\mathbf{A}^{\mathbf{N}}$, the relation $\Phi_A(a) = u$ is equivalent, according to formula (2), to the equalities

Let $u = (u_n)_{n \in \mathbf{N}}$ be in $\mathbf{A}^{\mathbf{N}}$. When $p \cdot 1_A$ is not a divisor of 0 in $\mathbf{A}$ (resp. when $p \cdot 1_A$ is invertible in $\mathbf{A}$), there exists at most one sequence $(a_n)_{n \in \mathbf{N}}$ in $\mathbf{A}$ (resp. exactly one sequence $(a_n)_{n \in \mathbf{N}}$ in $\mathbf{A}$) satisfying the equalities (10), whence a) and b).

Let us prove c). According to Lemma 2, the image $\mathbf{A}'$ of $\mathbf{A}^{\mathbf{N}}$ by $\Phi_A$ is the set of $u = (u_n)_{n \in \mathbf{N}}$ in $\mathbf{A}^{\mathbf{N}}$ such that $\sigma(u_n) \equiv u_{n+1} \mod. p^{n+1} \cdot \mathbf{A}$ for every $n \in \mathbf{N}$. It follows immediately that $\mathbf{A}'$ is a subring of $\mathbf{A}^{\mathbf{N}}$, stable under $f_A$ and $v_A$.