Proposition 3

Let A be a ring. a) The mapping $F_A$ is an endomorphism of the ring $W(A)$. b) The mapping $V_A$ is an endomorphism of the additive group underlying the ring $W(A)$. c) For every $a$ in $W(A)$, one has $F_A(V_A(a)) = p.a$ (sum in $W(A)$ of $p$ terms equal to $a$). d) Whatever $a$ and $b$ in $W(A)$ may be, one has

(sum in $W(A)$ of $p$ terms equal to $V_A(a \times b)$). e) Put $\mu = V_A(1) = (0, 1, 0, ...)$. For every $b$ in $W(A)$, one has

$^1$ The letter $F$ is the initial of the name Frobenius, and the letter $V$ that of the German word Verschiebung.

f) For every element $a$ of $W(A)$ denote by $a^{*p}$ the product in $W(A)$ of $p$ elements equal to $a$. Then one has

Let $\rho : B \to A$ be a homomorphism of rings satisfying the conditions of Lemma 3 of No. 4. Then $W(\rho) : W(B) \to W(A)$ is a surjective homomorphism of rings, and $\Phi_B : W(B) \to B^N$ is an injective homomorphism of rings. Moreover, $f_B : B^N \to B^N$ is a homomorphism of rings. According to the formulas (26) and (28), one has

whence immediately assertion $a$. Assertion $b$ follows in an analogous manner from the formulas (27) and (29) and from the fact that $v_B$ is an endomorphism of the additive group underlying $B^N$.

Let $a$ be an element of $W(A)$, and choose an element $x$ of $W(B)$ which $W(\rho)$ maps onto $a$. Put $\xi = \Phi_B(x)$. It follows immediately from the definitions of $f_B$ and $v_B$ that one has $f_B(v_B(\xi)) = p.\xi$ (sum in $B^N$ of $p$ terms equal to $\xi$). According to formulas (26) and (27) (where A is replaced by B), the elements $F_B(V_B(x))$ and $p.x$ of $W(B)$ therefore have the same image $p.\xi$ by the injective mapping $\Phi_B$, and hence are equal. The formula $F_A(V_A(a)) = p.a$ then follows from relations (28) and (29). This proves $c$.

Reasoning in an analogous manner, we reduce the proof of formula (30) to that of the relation

for $\xi, \eta$ in $B^N$. Now this results from the equalities

Taking into account $b$ and $c$, formula (31) results from formula (30), where $b$ is replaced by $V_A(b)$. Formula (32) is the particular case $a = 1$ of formula (30).

In an analogous manner, we reduce the proof of formula (33) to that of the relation

where $\xi^p$ denotes the product in $B^N$ of $p$ elements equal to $\xi$. By Prop. 2, c) of No. 2, this is equivalent to the fact that for every $n \geqslant 0$, one has

Now, for every $n \geqslant 0$, one has, by loc. cit.,

since $\xi = \Phi_B(x)$; one deduces from this, thanks to Lemma 1 of No. 1,

This proves the desired relation.