The Stacks project

Remark 23.5.4. With $A \supset I$ as in Lemma 23.5.3, let $\delta : I \to I$ be a map satisfying (2) and (3) of the lemma. Then (1) holds too. First, we prove $\delta (nx) = n\delta (x) + \frac{n^ p - n}{p!}x^ p$ for $n \geq 1$ by induction on $n$. The case $n = 1$ is OK. Assume that it is OK for some $n$, then $\delta ((n + 1)x) = \delta (nx) + \delta (x) + (\sum _{i = 1}^{p - 1} \frac{n^ i}{i!(p - i)!})x^ p = (n + 1)\delta (x) + \left( \frac{n^ p - n}{p!} + \frac{(n + 1)^ p}{p!} - \frac{n^ p + 1}{p!} \right)x^ p = (n + 1)\delta (x) + \frac{(n + 1)^ p - (n + 1)}{p!}x^ p$, so it is also OK for $n + 1$. On the other hand, we have $\delta (nx) = n^ p\delta (x)$, hence $(n^ p - n)\delta (x) = \frac{n^ p - n}{p!}x^ p$. But for any $p$ there exists $n$ such that $p^2 \not\mid n^ p - n$. Namely, we have $(\mathbb {Z}/p^2\mathbb {Z})^\times \cong \mathbb {Z}/p(p-1)\mathbb {Z}$. So there exists $n$ such that $n^{p-1} \not\equiv 1 \mod p^2$. Hence we get $p!\delta (x)=x^ p$