If $X\longrightarrow S$, where $S=\operatorname{Spec}(k)$, is a proper smooth morphism of dimension $p$ between $\mathbb{F}\_{p}$ schemes with Frobenius $F$, then there is an exact pairing $\wedge:\Omega^{i}(M)\times\Omega^{n-i}\_{X/S}\longrightarrow\Omega^{n}\_{X/S}$.


I am currently reading Deligne and Illusie's paper (<https://eudml.org/doc/143480>), and I am stuck on page 255. The authors make the claim that the above-mentioned exact pairing induces another one
\begin{equation\*}
F\_{\*}\Omega^{i}\_{X/S}\times F\_{\*}\Omega^{p-i}\_{X/S}\longrightarrow F\_{\*}\Omega^{p}\_{X/S}\longrightarrow H^{p}F\_{\*}\Omega^{\bullet}\_{X/S}\longrightarrow\Omega^{p}\_{X/S}
\end{equation\*}
They make the claim that the last morphism (the inverse of which they construct a few paragraphs earlier) "is just the trace morphism" and that the exactness of the pairing somehow follows from this claim. I don't quite get what they mean by that and why that pairing is exact. Can somebody please explain it to me in as much detail as possible?



Degeneration of Hodge-de Rham spectral sequence, exactness of a pairing and the trace morphism
First of all, Deligne-Illusie worked with relative Frobenius $F = F\_{X/S}: X \to X'$, so the target of the composition should be $\Omega\_{X'/S}^p$ instead of $\Omega\_{X/S}^p$.


Your question about why the composition
$$F\_\*\Omega\_{X/S}^p \to H^pF\_\*\Omega\_{X/S}^\bullet \to \Omega\_{X'/S}^p$$
is the trace map is explained in detail in [Brion-Kumar's book](https://www-fourier.univ-grenoble-alpes.fr/%7Embrion/fbook.pdf); see Chapter 1.3, especially Lemma 1.3.6, which is the local expression of that composition. (The statements in Brion-Kumar are about absolute Frobenius, but it suffices to transport the $k$-structures under the Frobenius to obtain statements about $X'/k$.)


Once we know that the above composition is the trace map, we use the Grothendieck duality
to conclude that the pairing is perfect. It works as follows.


I'll use the notations $\omega\_{X/k} = \Omega\_{X/S}^p$ and $\omega\_{X'/k} = \Omega\_{X'/S}^p$. Since $F$ is finite, we have
$$F^!\omega\_{X'/k} = \omega\_{X/k}.$$
The perfect pairing $\Omega\_{X/S}^i \otimes \Omega\_{X/S}^{p-i} \to \omega\_{X/k}$ defines an isomorphism
$$\Omega\_{X/S}^i \simeq R\mathcal{Hom}(\Omega\_{X/S}^{p-i},\omega\_{X/k}) \simeq R\mathcal{Hom}(\Omega\_{X/S}^{p-i},F^!\omega\_{X'/k}).$$
Applying $RF\_\*$ together with the Grothendieck duality gives
$$RF\_\*\Omega\_{X/S}^i \simeq RF\_\*R\mathcal{Hom}(\Omega\_{X/S}^{p-i},F^!\omega\_{X'/k}) \simeq 
R\mathcal{Hom}(RF\_\*\Omega\_{X/S}^{p-i},\omega\_{X'/k}).$$
so
$$F\_\*\Omega\_{X/S}^i \simeq \mathcal{Hom}(F\_\*\Omega\_{X/S}^{p-i},\omega\_{X'/k})$$
because $F$ is finite. This defines a perfect pairing $F\_\*\Omega\_{X/S}^i \otimes F\_\*\Omega\_{X/S}^{p-i} \to \omega\_{X'/k}$ which is identical to the pairing in question by constrction.


