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Let $(R,m,k)$ be a discrete valuation ring, where $k=R/m$. Let $X\rightarrow \mathrm{Spec}R$ be a projective, integral and flat $R$-scheme. Let $\mathscr F$ be a coherent sheaf such that $H^i(X,\mathscr F)$ is a torsion for $i>0$, i.e. the $i$-th cohomology of $\mathscr F$ on the generic fiber vanishes. In this case, can we compare the length of $H^0(X,\mathscr F)$ and the dimension of $H^i(X_k,\mathscr F_k)$?

I notice that if we let $\mathscr I$ be the ideal sheaf of $X_k$, and assume $\mathscr F$ is locally free, then we have the exact sequence $$0\rightarrow \mathscr F\otimes \mathscr I\rightarrow \mathscr F\rightarrow \mathscr F|_{X_k}\rightarrow 0,$$ Applying cohomology, we get $$\cdots\rightarrow H^{i}(X,\mathscr F\otimes\mathscr I)\rightarrow H^i(X,\mathscr F)\rightarrow H^i(X,\mathscr F|_{X_k})\rightarrow H^{i+1}(X,\mathscr F\otimes \mathscr I)\rightarrow\cdots$$ But I am not sure is there any thing interesting about $H^{i}(X,\mathscr F\otimes\mathscr I)$ in this case.

The reason why I consider this problem is that if $\mathscr F$ has support in $X_k$, then we have the isomorphism $H^i(X,\mathscr F)\simeq H^i(X_k, \mathscr F|_{X_k})$, so I was wondering how is the slightly general case.

Any help is appreciated.

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  • $\begingroup$ Actually your $\mathscr{I}$ is the pull back of $m=(\pi )$, where $\pi $ is a unifomizing parameter for $R$. Thus the left hand map on your exact sequence is just $\ \times \pi : H^i(X,\mathscr{F})\rightarrow H^i(X,\mathscr{F})$. But I don't think you can say anything a priori on its kernel and cokernel. $\endgroup$
    – abx
    Commented Dec 3, 2021 at 8:25
  • $\begingroup$ Since $H^i(X, \mathcal{F})$ is only annihilated by some power of $\mathfrak{m}$, it does not have to be a $k$-vector space. $\endgroup$ Commented Dec 3, 2021 at 8:58
  • $\begingroup$ @PiotrAchinger, That's very true. I have edited the problem. $\endgroup$ Commented Dec 3, 2021 at 13:34
  • $\begingroup$ @abx, Well, you may put some suitable conditions on $\mathscr F$, like $\mathscr F$ is supported on $X_k$. $\endgroup$ Commented Dec 3, 2021 at 13:43

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