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Let $f:Xto operatornameSpecmathbfZ_p$ be a smooth proper $mathbfZ_p$-scheme and $F$ a coherent sheaf on $X$ which is flat over $mathbfZ_p$. Further suppose that $H^1(X_p, F_p)=0$ where $X_p$ denotes the special fiber $Xotimes_mathbfZ_p mathbfF_p$ of $X$ and $F_p$ the pullback of $F$ to $X_p$.

Question 1: Is it then true that $R^1f_*F$ has empty support?

Question 2: What has this to do with the torsion of (the global sections) of $F$? Can we conclude from this that $H^0(X,F)$ is a free $mathbfZ_p$-module?

This seems all to follow from statements in Chapter III, Section 12 of Hartshorne but I fail to understand what the meaning of the higher direct images $R^if_*F$ really is.

EDIT: I also happen to now that there is a short exact sequence

$0to H^0(X,F')to H^0(X,F'')to H^0(X,F)to 0$

and that the first two terms commute with base change by $mathbfF_p$, i.e.

$H^0(X,F')otimes F_p = H^0(X_p,F'_p)$ and the same is true for $F''$.

The proof that I am trying to understand is the proof of Lemma 1.1 here.

$newcommandSpecmathrmSpec$$newcommandZmathbbZ$$newcommandQmathbbQ$I don't have Hartshorne on me, but I assume that the cited section is on proper base change/semi-continuity.

Since $f:Xto Spec(Z_p)$ is proper, $mathcalF$ is $Z_p$-flat and coherent, and $S$ and $X$ are Noetherian then we know that the map $f_i:Spec(Z_p)to Z$ given by $smapsto dim_k(s)H^i(X_k(s),mathcalF_k(s))$ is upper semi-continuous for all $igeqslant 0$. In particular since $f_1$ is upper semi-continuous we know that $f^-1((-infty,1))$ is an open subset of $Spec(mathbbZ_p)$. Since you're assuming that $pin f^-1((-1,infty))$ you can deduce that $Spec(Z_p)=f^-1((-infty,1))$ since the only open subset of $Spec(Z_p)$ containing $p$ is $Spec(Z_p)$. Thus, $H^1(X_Q_p,mathcalF_Q_p)=0$.

Now, since $Spec(Z_p)$ is connected and reduced we can use the same theory to show that since $f_1$ is constant that $R^1 f_astmathcalF$ is locally free and, moreover, that we have an isomorphism $(R^1 f_astmathcalF)_k(s)cong H^1(X_k(s),mathcalF_k(s))=0$ for all $s$ so that $R^1 f_astmathcalF=0$.

EDIT: To answer your other question, it is true that $H^0(X,mathcalF)$ is a free $Z_p$-module. This is more sophisticated though (I don't think it's in Hartshorne?) but it's a classical result that follows from things written in Mumford's book on abelian varieties. I found a reference here in the form of Corollary 2.3.

Indeed, since $Spec(Z_p)$ is affine we know that $f_ast mathcalF=widetildeH^0(X,mathcalF)$ (see the second edit below) and so the claim you want to prove is that $f_ast mathcalF$ is locally free. This is then what's stated in the cited notes of Osserman.

EDIT EDIT: Just to point out that in your response to "I don't understand what $R^i f_astmathcalF$ is really" note that if you have a reasonable map (qcqs probably?) $f:XtoSpec(A)$ then $H^i(X,mathcalF)$ is an $A$-module and $R^if_astmathcalF$ is just the quasi-coherent sheaf $widetildeH^i(X,mathcalF)$. All of the above is then by way of asking questions like "Is it true that if $H^1(X,mathcalF)otimes_mathbbZ_pmathbbF_p=0$ that $H^1(X,mathcalF)=0$? Moreover, in this case, is it true then that $H^0(X,mathcalF)$ is a free $mathbbZ_p$-module?" The answers to both of these questions being yes by the above.