The Stacks project

Lemma 50.19.5. Let $p$ be a prime number. Let $X \to S$ be a smooth morphism of relative dimension $d$ of schemes in characteristic $p$. The relative Frobenius $F_{X/S} : X \to X^{(p)}$ of $X/S$ (Varieties, Definition 33.36.4) is finite syntomic and the corresponding map

\[ \Theta _{X/X^{(p)}} : F_{X/S, *}\Omega ^\bullet _{X/S} \to \Omega ^\bullet _{X^{(p)}/S} \]

is zero in all degrees except in degree $d$ where it defines a surjection.

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 37.59.10 the morphism $F_{X/S}$ is a local complete intersection morphism. Hence $F_{X/S}$ is finite syntomic (see More on Morphisms, Lemma 37.59.8).

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 √Čtale 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)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FW2. Beware of the difference between the letter 'O' and the digit '0'.