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Let $f$ be some homogenous polynomial of $d$.

Let $X = \operatorname{Proj} (k[x,y,z]/(f))$ where $k$ is algebraically closed field of characteristic $p >0$.

Now let $R$ be a $k$-algebra.

What is the relationship between $H^1_{fl}(X, \alpha_p)$ and $H^1_{fl}(X \times_{spec(k)} spec(R), \alpha_p)$? Can we compute $H^1_{fl}(X \times_{spec(k)} spec(R), \alpha_p)$ from $H^1_{fl}(X, \alpha_p)$ ( and some other information)?

More specificaly when is it true that $H^1_{fl}(X \times_{spec(k)} spec(R), \alpha_p) = H^1_{fl}(X, \alpha_p) \otimes_{k} R $ ?

Also by $\alpha_p$ I mean the following sheaf $\alpha_p(U) = \{ x \in \Gamma(U, \mathcal{O}_u): u^p =0 \} $

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    $\begingroup$ Because $\alpha_p$ is not a quasi-coherent sheaf, statements like this will rarely be true. For example, if $X = \mathbb P^1$ (e.g. $f = z$) and $R = k[t]$, then $H^1_{\text{fl}}(X,\alpha_p)$ is $0$, whereas $H^1_{\text{fl}}(X \times \operatorname{Spec} R, \alpha_p) = k[t]/k[t^p]$. So even if $H^1_{\text{fl}}(X,\alpha_p)$ is zero, this need not remain true for $H^1_{\text{fl}}(X \times \operatorname{Spec} R, \alpha_p)$. This example probably shows that there is no 'functorial' way to compute $H^1_{\text{fl}}(X \times \operatorname{Spec} R, \alpha_p)$ from $H^1_{\text{fl}}(X, \alpha_p)$. $\endgroup$
    – R. van Dobben de Bruyn
    Sep 23, 2018 at 22:05

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