Remark 33.36.11. Let $p > 0$ be a prime number. Let $S$ be a scheme in characteristic $p$. Let $X$ be a scheme over $S$. For $n \geq 1$

$X^{(p^ n)} = X^{(p^ n/S)} = X \times _{S, F_ S^ n} S$

viewed as a scheme over $S$. Observe that $X \mapsto X^{(p^ n)}$ is a functor. Applying Lemma 33.36.2 we see $F_{X/S, n} = (F_ X^ n, \text{id}_ S) : X \longrightarrow X^{(p^ n)}$ is a morphism over $S$ fitting into the commutative diagram

$\xymatrix{ X \ar[rr]_{F_{X/S, n}} \ar[rrd] \ar@/^1em/[rrrr]^{F_ X^ n} & & X^{(p^ n)} \ar[rr] \ar[d] & & X \ar[d] \\ & & S \ar[rr]^{F_ S^ n} & & S }$

where the right square is cartesian. The morphism $F_{X/S, n}$ is sometimes called the $n$-fold relative Frobenius morphism of $X/S$. This makes sense because we have the formula

$F_{X/S, n} = F_{X^{(p^{n - 1})}/S} \circ \ldots \circ F_{X^{(p)}/S} \circ F_{X/S}$

which shows that $F_{X/S, n}$ is the composition of $n$ relative Frobenii. Since we have

$F_{X^{(p^ m)}/S} = F_{X^{(p^{m - 1})}/S}^{(p)} = \ldots = F_{X/S}^{(p^ m)}$

(details omitted) we get also that

$F_{X/S, n} = F_{X/S}^{(p^{n - 1})} \circ \ldots \circ F_{X/S}^{(p)} \circ F_{X/S}$

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