Let $U$ be a open affine subscheme of a smooth, proper scheme $X$ over $\mathbf{Z}_p$. Over $\mathbf{Q}_p$ we know that $\mathrm{H}^i(U \times \mathbf{Q}_p/\mathbf{Q}_p)$ is finite-dimensional (where $\mathrm{H}^i(\cdot/R) = \mathbf{H}^i(\Omega_{\cdot/R}^\bullet)$ is the $i$-th de Rham cohomology of a scheme over a ring $R$). My question is:
Is the image of $\mathrm{H}^i(U/\mathbf{Z}_p)$ in $\mathrm{H}^i(U \times \mathbf{Q}_p/\mathbf{Q}_p)$ finitely generated as $\mathbf{Z}_p$-module?
I am particularly interested in the case of (smooth, geometrically intgeral) curves and $i=1$. For simplicity, let $Z = X \setminus U$ be a single $\mathbf{Z}_p$-rational point. Here are my partial results:
If $X$ is of genus $0$, then the answer is obviously 'yes' as $\mathrm{H}^1(U \times \mathbf{Q}_p/\mathbf{Q}_p)$ vanishes.
If $X$ is ordinary and of genus $1$, then the answer is 'no'.
The answer can also be 'no' if $X$ is of genus $1$ and supersingular. (An example is: $U$ defined by $y^2 = x^3 + 1$ over $\mathbf{Z}_5$. The calcuations are somewhat lengthy.)
Let me give you the starting point for the proof/example: The image of $\mathrm{H}^1(U/\mathbf{Z}_p)$ is finitely generated if and only if there exists an integer $\ell$ such that $\mathrm{H}^0(\Omega_X(\ell Z))$ surjects onto the image. This happens if and only if for every differential $\omega \in \mathrm{H}^0(\Omega_U)$ there exists a function $f \in \mathrm{H}^0({\cal{O}}_U)$ such that $p^r\omega - df \in p^r\mathrm{H}^0(\Omega_X(\ell Z))$. (Note that reducing to $\mathrm{H}^0(\Omega_X(\ell Z))$ is not enough.) In particular, one must be able to reduce differentials with Laurent tail $t^{-p^r-1}dt + O(t^{-p^r})dt$ (The function $t$ is such that $\{ t, p \}$ are local coordinates for the stalk of ${\cal{O}}_X$ at the special point of $Z$. In particular, $t$ is a local uniformizer over $\mathbf{Q}_p$ as well as over $\mathbf{F}_p$.) This is only possible if the following holds:
For every $r \in \mathbf{N}$, is there a function $f \in \mathrm{H}^0({\cal{O}}_X(p^rZ))$ with Laurent tail $t^{-p^r} + O(t^{-p^r +1})$ and $df \in p^r\mathrm{H}^0(\Omega_X((p^r+1)Z))$?
Update: If $X$ is ordinary of genus $1$, here's how to show that such a function cannot exist. The main ingredients are reduction modulo $p$ and the residue theorem. A $\bar{\cdot}$ will denote reduction modulo $p$ of the object in question. Since $p^r$ divides $df$, the Laurent expansion of $\bar{f}$ with respect to $\bar{t}$ is $\bar{t}^{-p^r} + O(\bar{t})$ (we can always assume that the coefficient of $\bar{t}^0$ is $0$). Choose a generator $\bar{\vartheta}$ of $\mathrm{H}^0(\Omega_{\bar{X}})$ and let $\sum_{i \ge 0}\bar{\vartheta}_i\bar{t}^id\bar{t}$ be its Laurent expansion. By the residue theorem $0 = \mathrm{res}(\bar{f}\bar{\vartheta}) = \mathrm{res}_{\bar{Z}}(\bar{f}\bar{\vartheta}) = \bar{\vartheta}_{p^r-1}$ (all other summands contributing to the residue are $0$). But $\bar{\vartheta}_{p^r-1} = \bar{c}\bar{\vartheta}_0 = \bar{c}$ (the trace of the Frobenius modulo $p$, see [1]). Since $X$ is ordinary, $\bar{c}$ cannot be $0$. This is the contradiction.
[1] Lang, Elliptic functions, Appendix 2.