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Consider a smooth projective curve $X$ over an algebraically closed field $k$, and denote by $\omega_{X}$ its canonical line bundle.

Let $\mathcal X = X × S$ and let $q$ be the first projection with $X,S$ schemes.

Is it true that $f^*\Omega_{\mathcal X|S} = q^* \omega^p_{X}$ with $f$ the absolute frobenius of $\mathcal X$? how can we see it?

These are the exact notations taken from this paper , I guess I have to read: $f^*\Omega_{\mathcal X|S} = q^* \omega^{\otimes p}_{X}$?

Thank you.

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    $\begingroup$ what do you mean by "canonical line bundle" for an arbitrary scheme? I think on some schemes the canonical sheaf will not be a line bundle and on some schemes there is no canonical sheaf at all. $\endgroup$
    – user145520
    Mar 8, 2020 at 17:53
  • $\begingroup$ From hartshorne p $180$ the canonical sheaf of a $k$-scheme $X$ is $\omega_X=\bigwedge^n \Omega_{X|k}$ $\endgroup$
    – Conjecture
    Mar 8, 2020 at 20:22
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    $\begingroup$ So you mean that $X$ is of finite type over a field $k$? Please state your assumptions (in the original question). Also the tag Frobenius algebras is inappropriate. $\endgroup$ Mar 8, 2020 at 20:41
  • $\begingroup$ I just removed the tag, thank you. I updated the question with the assumption from the paper. I think of the caconical line bundle as a locally free $\mathcal O_X-$module of rank $1$ (i.e. an invertible sheaf). Since $X$ is a smooth curve (of dim $1$), I guess $\omega_X=\Omega_{X|X}$ which is invertible. $\endgroup$
    – Conjecture
    Mar 8, 2020 at 21:50

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