# pullback of Frobenius morphism and pull back of a projection

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.

• 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.
– user145520
Mar 8, 2020 at 17:53
• From hartshorne p $180$ the canonical sheaf of a $k$-scheme $X$ is $\omega_X=\bigwedge^n \Omega_{X|k}$ Mar 8, 2020 at 20:22
• 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. Mar 8, 2020 at 20:41
• 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. Mar 8, 2020 at 21:50