# Is de Rham cohomology of affine schemes over discrete valuation rings finitely generated (modulo torsion)?

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:

1. 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.

2. If $X$ is ordinary and of genus $1$, then the answer is 'no'.

3. 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.

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After more studying, I found a solution. An $\overline{\cdot}$ will denote the reduction modulo $p$ of the object in question. Let us assume that $\overline{X}$ is a geometrically integral curve of genus $g$ and that $Z = X \setminus U$ is a (relative, reduced) normal crossing divisor. Then the following holds:

If $\gamma$ is the Hasse-Witt invariant of $\overline{X}$, i.e. the rank of the $p$-torsion of the Jacobian of $\overline{X}$, then $$\mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion}) \cong \mathbf{Q}_p^{2g-\gamma} \oplus \mathbf{Z}_p^{\deg{\overline{Z}}-1+\gamma}.$$

The module $\mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion})$ embeds into $\mathrm{H}_{\mathrm{dR}}^1(X_{\mathbf{Q}_p} \setminus Z_{\mathbf{Q}_p})$ and this space is known to be isomorphic to $\mathbf{Q}_p^{2g + \deg{Z} - 1}$. Thus $\mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion})$ is isomorphic (as $\mathbf{Z}_p$-module) to $\mathbf{Q}_p^r \oplus \mathbf{Z}_p^s$ with $r+s = 2g + \deg{Z}-1$. Tensoring with $\mathbf{F}_p$ gives $\mathbf{F}_p^s$. Since $\deg{Z} = \deg{\overline{Z}}$, we need to show that the dimension of $\mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion}) \otimes \mathbf{F}_p$ is $\deg{\overline{Z}}-1+\gamma$.

Let $d{\cal{O}}_{X}(X \setminus Z) \colon (p^\infty)$ be the $p$-saturation of $d{\cal{O}}_{X}(X \setminus Z)$, i.e. it contains all differentials $\omega$ such that $p^r\omega = df$ for some $r \in \mathbf{N}$ and $f \in {\cal{O}}_X(X \setminus Z)$. Then there is the exact sequence $d{\cal{O}}_{X}(X \setminus Z) \colon (p^\infty) \to \Omega_X(X \setminus Z) \to \mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion}) \to 0$. Since tensoring with $\mathbf{F}_p$ is right-exact, we get $$d{\cal{O}}_{X}(X \setminus Z) \colon (p^\infty) \otimes \mathbf{F}_p \to \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z}) \to \mathrm{H}_{\mathrm{dR}}^1(X \setminus Z) / (\mathrm{torsion}) \otimes \mathbf{F}_p \to 0.$$

Write $\Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$ for the image of $d{\cal{O}}_{X}(X \setminus Z) \colon (p^\infty) \otimes \mathbf{F}_p$ in $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})$. Its elements are called pseudo-exact differentials [1, Sec. 2]. What we want to prove follows from the two claims

$$\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z}) \big/ \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) \cong \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \big/ \left( \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) \right)$$

and

$$\dim{\Omega_{\overline{X}}(\overline{Z})(\overline{X}) \big/ \left( \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) \right)} = \deg{\overline{Z}} - 1 + \gamma.$$

Concerning the second claim, note that pseudo-exact differentials are residue-free [1, Sec. 3] and that $\overline{Z}$ is reduced since $Z$ is a reduced normal crossing divisor. Thus, $\dim{\Omega_{\overline{X}}(\overline{Z})(\overline{X}) / ( \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})} )$ is the sum of $\dim{\Omega_{\overline{X}}(\overline{Z})(\overline{X}) / \Omega_{\overline{X}}(\overline{X})}$ and $\dim{\Omega_{\overline{X}}(\overline{X}) / \Omega_{\overline{X}}(\overline{X}) \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})}$. The former dimension is $\deg{\overline{Z}}-1$ and the latter is $\gamma$ [4, Cor. 1].

Let $C$ be the Cartier operator on the differentials of $\overline{X}$. If $\overline{x}$ is a separating element for the function field $K(\overline{X})$, then every differential may be written as $(\sum_{i=0}^{p-1} \lambda_i\overline{x}^i)d\overline{x}$ with $\lambda_i \in k$ and $C$ maps this differential to $\lambda_i^{1/p} d\overline{x}$. The Cartier operator restricts to a map (also called $C$) on $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})$ and [3, Sec. 1] $$\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z}) = \bigcup_{i \in \mathbf{N}} C^{-i}(\Omega_{\overline{X}}(\overline{Z})(\overline{X})) \hspace{10pt} \text{and} \hspace{10pt} \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) = \bigcup_{i \in \mathbf{N}} C^{-i}(\{ 0 \}).$$ Let $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} := C^{-i}(\Omega_{\overline{X}}(\overline{Z})(\overline{X}))$ and $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i, \mathrm{p-ex})} := \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$. Then $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} / \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i, \mathrm{p-ex})}$ is isomorphic to $\Omega_{\overline{X}}(\overline{Z})(\overline{X}) / \Omega_{\overline{X}}(\overline{Z})(\overline{X}) \cap \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$. But $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} / \Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i, \mathrm{p-ex})}$ is also isomorphic to $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z})^{(i)} + \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z}) / \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$. Varying $i$, these spaces form a direct system whose limit is $\Omega_{\overline{X}}(\overline{X} \setminus \overline{Z}) / \Omega_{\overline{X}}^{(\mathrm{p-ex})}(\overline{X} \setminus \overline{Z})$ and whose maps are isomorphisms. This finishes the proof. (Compare with [2, Satz 3].)

[1] Lamprecht "Zur Klassifikation von Differentialen in Körpern von Primzahlcharakteristik. I" 1958

[2] Kunz "Einige Anwendungen des Cartier-Operator" 1962

[3] Kodama "Residuenfreie Differentiale und der Cartier-Operator algebraischer Funktionenkörper" 1971

[4] Kodama "On the Rank of Hasse-Witt Matrix" 1984

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