Lemma 1

If x and y are two elements of A congruent modulo $\mathbf{J}_m$, one has

By induction on n, one reduces to the case where $n = 1$. Let P denote the polynomial $\sum_{i=0}^{p-1} X^i Y^{p-1-i}$ of $\mathbf{Z}[X, Y]$. In view of the assumption made on x and y, one has $P(x, y) \equiv P(x, x) \equiv p.x^{p-1} \mod \mathbf{J}_m$. Now one has $\mathbf{J}_m + p.\mathbf{A} \subset \mathbf{J}_1$, whence $P(x, y) \in \mathbf{J}_1$. Finally, $x^p - y^p = (x - y) P(x, y)$ belongs to $\mathbf{J}_m \mathbf{J}_1 \subset \mathbf{J}_{m+1}$.

Let us prove a) by induction on n. The case $n = 0$ is immediate. Suppose $n \geqslant 1$. Under the hypotheses of a), one has (4) $a_i^p \equiv b_i^p \mod \mathbf{J}_{m+1}$ for $0 \leqslant i \leqslant n-1$ according to Lemma 1 , (5) $\Phi_{n-1}(a_0^p, ..., a_{n-1}^p) \equiv \Phi_{n-1}(b_0^p, ..., b_{n-1}^p) \mod \mathbf{J}_{m+n}$ according to the induction hypothesis applied to the elements $a_0^p, ..., a_{n-1}^p, b_0^p, ..., b_{n-1}^p$ of A, and (6) $\Phi_n(a_0, ..., a_n) - p^n.a_n \equiv \Phi_n(b_0, ..., b_n) - p^n.b_n \mod \mathbf{J}_{m+n}$ according to formulas (2) and (5). Since $a_n - b_n$ belongs to $\mathbf{J}_m$, the element $p^n.a_n - p^n.b_n$ belongs to $\mathbf{J}_{m+n}$ and one deduces from (6) the congruence

whence a).

Let us prove b) by induction on n. The case $n = 0$ is immediate. Suppose $n \geqslant 1$. Under the assumptions of b), we have $a_i \equiv b_i \mod \mathbf{J}_m$ for $0 \leqslant i \leqslant n-1$ by the induction hypothesis, and we deduce from this, as previously, the congruences (4), (5) and (6). But by assumption $\Phi_n(a_0, ..., a_n)$ and $\Phi_n(b_0, ..., b_n)$ are congruent mod. $\mathbf{J}_{m+n}$, and hence we have $p^n.(a_n - b_n) \in \mathbf{J}_{m+n}$. Since the relation $p.x \in \mathbf{J}_{k+1}$ implies $x \in \mathbf{J}_k$ for every $x \in \mathbf{A}$ and every $k \geqslant 1$, we have $a_n - b_n \in \mathbf{J}_m$, which completes the proof.