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But if I am not misunderstanding, you mean $H_0$ and $H_1$ (which are vector bundles) locally free but not $H^0$ and $H^1$ (which are cohomology functor).
Thank you!So by the same arguement, if we hava a $Z_p$ algebra $\bar{R}$, such that $\bar{R}/p\bar{R}$ is perfect, then we will have $\hat{\bar{R}}=W(\bar{R}/p\bar{R})$, is that right ?