I am reading the survey paper: "The de-Rham Witt complex and Crystalline cohomology" by Luc Illusie.


In math line (2.1.12), Illusie considers the pairing $\langle-,-\rangle:\Omega\_{X/S}^1\times T\_{X/S}\longrightarrow \mathcal{O}\_X$ of the tangent and cotangent bundles on a scheme $X$ of characteristic $p$, relative to a morphism $X\longrightarrow S$. ($S$ also has characteristic $p$).


Earlier we defined a bunch of maps, which would be necessary to describe in order for me to ask my question. The first one is the Cartier operation, $C$, which sends a closed form $w\in \Omega\_{X/S}^1$ to a 1-form $Cw \in \Omega\_{X^{(p)}/S}^1$, where $X^{(p)}$ is the base change of $X\longrightarrow S$ with respect to the absolute Frobenius on $S$. Denote by $W$ the canonical projection $W:X^{(p)}\longrightarrow X$.


Illusie then wants to say something about the pairing
$$\langle Cw,W^\*D\rangle$$
which happens on $\Omega\_{X^{(p)}/S}^1\times T\_{X^{(p)}/S}\longrightarrow \mathcal{O}\_{X^{(p)}}$. Namely, he gives the identity
$$\langle Cw,W^\*D\rangle^p = \langle w,D^p\rangle - D^{p-1}\langle w,D\rangle.$$
My questions are:


1. What does the $p$-th power of a tangent vector, $D^p$, mean? Does the tangent space have a ring structure?
2. The pairing apriori should have values in $\mathcal{O}\_{X^{(p)}}$, once it is raised to the $p$'th power we regard it as having values in $\mathcal{O}\_X$ as this is. For this reason it seems that Illusie is regarding $D^{p-1}$ as an element of $\mathcal{O}\_X$, which is consistent with math line (2.1.13), why is that? I guess this goes back to the first question.
3. Illusie also writes $D\_i^p = 0$, where $D\_i = \partial/\partial x\_i$, for some etale basis $(x\_i)$, why is that true? I guess that this should all be clear once I understand this notation.


Thanks in advance!



Pairing of cotangent and tangent bundles
For (1), recall that if $R$ is a ring, then a derivation $D: R \to R$ satisfies the Leibniz rule, which by induction on $n$ implies that if $D^n$ denotes the $n$-fold iterate of $D$, then
$$D^n(fg) = \sum\_{i=0}^n \binom{n}{i} D^i(f) D^{n-i}(g).$$
Since $\binom{p}{i} \equiv 0$ for $0<i<p$, this implies that if $R$ is an $\mathbf{F}\_p$-algebra, then $D^p(fg) = D^p(f) g + f D^p(g)$. In other words, $D^p$ is a derivation.


Let me answer (3) before (2). If you work locally, i.e., consider the derivation $\partial\_x$ on $\mathbf{F}\_p[x]$, then $(\partial\_x)^p x^n$ is zero for $n<p$, and is $p! \binom{n}{p} x^{n-p}$ for $n\geq p$, which is zero. So the derivation $(\partial\_x)^p$ is identically zero.


Let's now discuss (2); since everything is local on $X$ and $X$ is smooth, we can assume that $X$ is etale over $\mathbf{A}^n\_S$ with basis $x\_1, \cdots, x\_n$. Let $\omega$ be a closed $1$-form on $X$. Because of the Cartier isomorphism $\mathfrak{C}: \mathcal{H}^i(F\_\ast \Omega^\bullet\_{X/S}) \xrightarrow{\sim} \Omega^i\_{X^{(p)}/S}$, we can write $\omega = df + \sum\_{i=1}^n F^\ast(g\_i) x\_i^{p-1} dx\_i$ for some functions $g\_i$ and $f$ on $X$. Both $\langle \mathfrak{C} \omega, D\rangle^p$ and $\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle$ kill $df$, so by Frobenius semilinearity, we can assume that $\omega = x\_i^{p-1} dx\_i$. We'll also just assume $n=1$ and write $x$ instead of $x\_1$.


Then $\langle \mathfrak{C} \omega, D\rangle = \langle dx^{(p)}, D\rangle = D(x)$, and $\langle \omega, D^p\rangle - D^{p-1} \langle \omega, D\rangle = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx)$. So we need to prove that
$$(Dx)^p = x^{p-1} D^p(x) - D^{p-1}(x^{p-1} Dx),$$
which is an identity due to Hochschild. See <https://joshuamundinger.github.io/assets/notes/hochschild-identity.pdf> for a cute argument; it reduces to using the multinomial analogue of the Leibniz rule for $D^p(x^p)$ and a multinomial analogue of the binomial coefficient vanishing.




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Clarification from comments: let's again just consider a single variable $x$ and assume $X = \mathbf{A}^1\_{\mathbf{F}\_p}$. The most general closed $1$-form is $\omega = df + g(x^p) x^{p-1} dx$. Then $\mathfrak{C}(df) = 0$, so we can take $a\_i$ to be $g(x^p) x^{p-1}$. Now, $\partial\_x$ is $\mathbf{F}\_p[x^p]$-linear, so
$$\partial\_x^{p-1} (g(x^p) x^{p-1}) = g(x^p) \cdot (\partial\_x^{p-1} x^{p-1}) = (p-1)! g(x^p) = -g(x^p).$$
This means that $\mathfrak{C}(g(x^p) x^{p-1} dx) = g(x^{(p)}) dx^{(p)}$ can be written as $-\partial\_x^{p-1} (g(x^p) x^{p-1}) dx^{(p)}$, as desired.


