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Let $X$ be a smooth projective variety over an algebraically closed field $k$.

When $k= \mathbb C$, it is known that $h^{1,1}(X)=h^1(X,\Omega_X) > 0$. Is this true also when $\rm char(k)=p > 0$?

Is the natural map $\rm Pic(X)=H^1(X,\mathcal O_X^*) \to H^1(X,\Omega_X)$ non-zero?

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    $\begingroup$ Yes to both questions. See for instance A. Grothendieck, La théorie des classes de Chern, Bull. SMF 86 (1958), p. 137-154, where he constructs (among many other things) a theory of Chern classes with values in the Hodge cohomology $\ \bigoplus H^p(X,\Omega ^p_X)$. $\endgroup$
    – abx
    Oct 25, 2022 at 13:01
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    $\begingroup$ @abx: thanks for the reference. Now how to prove that the map $\rm Pic(X) \to H^1(\Omega^1_X)$ is non-zero? I suppose one takes an ample line bundle $L$ on $X$ and then shows that $c_1(L) \in H^1(\Omega^1_X)$ is non-zero. But with this axiomatic definition of $c_1$ this is not clear to me. May be using intersection theory? $\endgroup$
    – Cob
    Oct 25, 2022 at 14:36
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    $\begingroup$ I think I got it: by definition, $c_1(L)=-\xi_L$. Now, via the isomorphism $\mathbb P(L) \to X$ we have that $\xi_L=[L]$, hence is non-zero. $\endgroup$
    – Cob
    Oct 25, 2022 at 15:06
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    $\begingroup$ Not clear to me either, and I didn't find it in the reference (though I didn't read too carefully). I am not sure I believe that the map should always be nonzero. Your argument above cannot work: replace $L$ with $L^p$, then $c_1(L^p) = pc_1(L) = 0$. So of course it sufficient if there exists a curve class $C$ and a line bundle $L=O(D)$ such that $(C.D)$ is coprime to $p$. But, take a surface $X$ with ${\rm Pic}(X)=\mathbf{Z}\cdot [L]$ where $L=O(D)$ and $D^2$ is divisible by $p$... $\endgroup$
    – Piotr Achinger
    Oct 25, 2022 at 15:29
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    $\begingroup$ @Piotr Achinger: got it, my argument is not ok, thanks. $\endgroup$
    – Cob
    Oct 25, 2022 at 16:17

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