Lemma 1

Let $p$ be a prime number and $i$ an integer in the range from 1 to $p - 1$, then the binomial coefficient $\binom{p}{i}$ is an integer divisible by $p$.

We argue by induction on $i$, the case $i = 1$ being immediate from the formula $\binom{p}{1} = p$. Suppose that $2 \leq i \leq p - 1$ and that $\binom{i}{i-1}$ is divisible by $p$. Then the integer $i \binom{p}{i} = (p - i + 1) \binom{p}{i-1}$ belongs to the prime ideal $p\mathbf{Z}$ of $\mathbf{Z}$; since $i \notin p\mathbf{Z}$, we have $\binom{p}{i} \in p\mathbf{Z}$ and the lemma follows.

Let $A$ be a commutative ring of characteristic $p$ and $f$ an integer $\geq 0$. From Th. 2 we deduce by induction on $f$ that the mapping $a \mapsto a^{p^f}$ is an endomorphism of the ring $A$. In particular we have the relation

for any $a_1, \ldots, a_n$ in $A$. The mapping $a \mapsto a^p$ is sometimes called the Frobenius endomorphism of $A$. Taking $A = \mathbf{F}_p$ and $a_i = 1$, we obtain from (3) the relation:

For each subset $S$ of $A$ we denote by $S^{p^f}$ the set of elements of $A$ of the form $x^{p^f}$ with $x \in S$^1. In particular, if $K$ is a subring of $A$, the set $K^{p^f}$ is a subring of $A$. If $K$ is a subring of $A$ and $S$ a subset of $A$ we denote by $K[S]$ the subring of $A$ generated by $K \cup S$; when $A$ is a field, we denote by $K(S)$ the field of fractions of $K[S]$, that is, the subfield of $A$ generated by $K \cup S$.