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Let $L|F$ be a extension of perfect fields of characteristic $p$, $\phi_F:F \to F_{\phi}$, $\phi_L:L \to L_{\phi}$ the Frobenius isomorphisms ($F_{\phi}=F$ but considered as $F$-algebra via $\phi_F$). The induced homomorphisms on the differential modules $\Omega_{L|K}\otimes_LL_{\phi} \to \Omega_{L_{\phi}|K}$ and $\Omega_{L_{\phi}|K} \to \Omega_{L_{\phi}|K_{\phi}}$ are then isomorphisms, but their composition $\Omega_{L|K}\otimes_LL_{\phi} \to \Omega_{L_{\phi}|K_{\phi}}$ takes $dX$ to $dX^p=pX^{p-1}dX = 0$ and then $\Omega_{L|K}=0$.

I find it a little strange (when $L|F$ is transcendental). Am I wrong?

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  • $\begingroup$ My impression is that the indicated maps from the tensor products do not exist. Am I wrong? $\endgroup$
    – ACL
    Commented Sep 30, 2014 at 21:49
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    $\begingroup$ If $L$ is perfect then every $x \in L$ has the form $x = y^p$ for some $y \in L$, so ${\rm{d}}x = 0$ in $\Omega_{L/K}$ for every subfield $K$ of $L$ without needing that blizzard of other mappings (and hence $\Omega_{L/K}$ always vanishes). What makes the vanishing seem strange? $\endgroup$
    – user27920
    Commented Sep 30, 2014 at 23:40

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