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
Comments (0)