**Proof.**
Observe that $F_{X/S}$ is a finite morphism by Varieties, Lemma 33.36.8. To prove that $F_{X/S}$ is flat, it suffices to show that the morphism $F_{X/S, s} : X_ s \to X^{(p)}_ s$ between fibres is flat for all $s \in S$, see More on Morphisms, Theorem 37.16.2. Flatness of $X_ s \to X^{(p)}_ s$ follows from Algebra, Lemma 10.128.1 (and the finiteness already shown). By More on Morphisms, Lemma 37.62.10 the morphism $F_{X/S}$ is a local complete intersection morphism. Hence $F_{X/S}$ is finite syntomic (see More on Morphisms, Lemma 37.62.8).

For every point $x \in X$ we may choose a commutative diagram

\[ \xymatrix{ X \ar[d] & U \ar[l] \ar[d]_\pi \\ S & \mathbf{A}^ d_ S \ar[l] } \]

where $\pi $ is étale and $x \in U$ is open in $X$, see Morphisms, Lemma 29.36.20. Observe that $\mathbf{A}^ d_ S \to \mathbf{A}^ d_ S$, $(x_1, \ldots , x_ d) \mapsto (x_1^ p, \ldots , x_ d^ p)$ is the relative Frobenius for $\mathcal{A}^ d_ S$ over $S$. The commutative diagram

\[ \xymatrix{ U \ar[d]_\pi \ar[r]_{F_{X/S}} & U^{(p)} \ar[d]^{\pi ^{(p)}} \\ \mathbf{A}^ d_ S \ar[r]^{x_ i \mapsto x_ i^ p} & \mathbf{A}^ d_ S } \]

of Varieties, Lemma 33.36.5 for $\pi : U \to \mathbf{A}^ d_ S$ is cartesian by Étale Morphisms, Lemma 41.14.3. Since the construction of $\Theta $ is compatible with base change and since $\Omega _{U/S} = \pi ^*\Omega _{\mathbf{A}^ d_ S/S}$ (Lemma 50.2.2) we conclude that it suffices to show the lemma for $\mathbf{A}^ d_ S$.

Let $A$ be a ring of characteristic $p$. Consider the unique $A$-algebra homomorphism $A[y_1, \ldots , y_ d] \to A[x_1, \ldots , x_ d]$ sending $y_ i$ to $x_ i^ p$. The arguments above reduce us to computing the map

\[ \Theta ^ i : \Omega ^ i_{A[x_1, \ldots , x_ d]/A} \to \Omega ^ i_{A[y_1, \ldots , y_ d]/A} \]

We urge the reader to do the computation in this case for themselves. As in Example 50.19.4 we may reduce this to computing a formula for $\Theta ^ i$ in the universal case

\[ \mathbf{Z}[y_1, \ldots , y_ d] \to \mathbf{Z}[x_1, \ldots , x_ d],\quad y_ i \mapsto x_ i^ p \]

In turn, we can find the formula for $\Theta ^ i$ by computing in the complex case, i.e., for the $\mathbf{C}$-algebra map

\[ \mathbf{C}[y_1, \ldots , y_ d] \to \mathbf{C}[x_1, \ldots , x_ d],\quad y_ i \mapsto x_ i^ p \]

We may even invert $x_1, \ldots , x_ d$ and $y_1, \ldots , y_ d$. In this case, we have $\text{d}x_ i = p^{-1} x_ i^{- p + 1}\text{d}y_ i$. Hence we see that

\begin{align*} \Theta ^ i( x_1^{e_1} \ldots x_ d^{e_ d} \text{d}x_1 \wedge \ldots \wedge \text{d}x_ i) & = p^{-i} \Theta ^ i( x_1^{e_1 - p + 1} \ldots x_ i^{e_ i - p + 1} x_{i + 1}^{e_{i + 1}} \ldots x_ d^{e_ d} \text{d}y_1 \wedge \ldots \wedge \text{d}y_ i ) \\ & = p^{-i} \text{Trace}(x_1^{e_1 - p + 1} \ldots x_ i^{e_ i - p + 1} x_{i + 1}^{e_{i + 1}} \ldots x_ d^{e_ d}) \text{d}y_1 \wedge \ldots \wedge \text{d}y_ i \end{align*}

by the properties of $\Theta ^ i$. An elementary computation shows that the trace in the expression above is zero unless $e_1, \ldots , e_ i$ are congruent to $-1$ modulo $p$ and $e_{i + 1}, \ldots , e_ d$ are divisible by $p$. Moreover, in this case we obtain

\[ p^{d - i} y_1^{(e_1 - p + 1)/p} \ldots y_ i^{(e_ i - p + 1)/p} y_{i + 1}^{e_{i + 1}/p} \ldots y_ d^{e_ d/p} \text{d}y_1 \wedge \ldots \wedge \text{d}y_ i \]

We conclude that we get zero in characteristic $p$ unless $d = i$ and in this case we get every possible $d$-form.
$\square$

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