Exercise 38, § 1

Let $p$ be a prime number, A a ring, and $a \in U(A)$. a) We have $F_p(a) \equiv a^p \mod. pA^\mathbf{J}$; in other words, we have $F_{p,n}(a) \equiv a_n^p \mod. pA$ for every $n \in \mathbf{J}$. b) We have $F_p(a) \equiv a^{*p} \mod. p \times U(A)$, where $a^{*p}$ denotes the product in U(A) of $p$ terms equal to $a$, and where $p \times U(A)$ denotes the ideal of U(A) generated by $p \times 1_{U(A)}$, the sum in U(A) of $p$ terms equal to $1_{U(A)}$. (It is enough to treat the case where $A = \mathbf{Z}[X]$ and $a = X$. Then A has no $\mathbf{Z}$-torsion and $\Phi : U(A) \to A^\mathbf{J}$ is injective. For a) it is enough (p. 51, Exerc. 30, b)) to prove that, for every $n \in \mathbf{J}$, $\Phi_n(F_p(X)) = \Phi_{pn}(X)$ is congruent to $\Phi_n(X^p)$ modulo $p^{w_p(pn)}A$, which results from Exerc. 29, p. 51. For b) it is enough to prove that $\Phi(F_p(X)) = f_p(\Phi(X))$ is congruent to $\Phi(X^{*p}) = \Phi(X)^p$ modulo $p.\Phi(U(A))$. Now there exists an element $u \in A^\mathbf{J}$ such that

and it is a question of proving that $u \in \Phi(U(A))$. For every prime number $q$, let $\sigma_q$ be the endomorphism of the ring $A = \mathbf{Z}[X]$ such that $\sigma_q(X_n) = X_n^q$ for every $n \in \mathbf{J}$. By Exerc. 32, p. 51, it is enough to prove that, for every prime number $q$ and for every $n \in \mathbf{J}$, we have $u_{qn} \equiv \sigma_q(u_n) \mod. q^{w_q(qn)}A$, which is equivalent to proving the following congruence (1) Now the right-hand term is equal to $\Phi_{pn}(X^q) - \Phi_n(X^q)^p$. By Exerc. 29, p. 51, we have (2) and (3) The right-hand and left-hand terms of (1) are both congruent to zero modulo $pA$ (p. 51, Exerc. 29), hence (1) results from (2) and (3) if $q \neq p$. Suppose that we have $q = p$. Then it results from (3) and Lemma 1 that we have (4) and (1) results from (2) and (4).)