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Let $k$ be a perfect field of positive characteristic and denote by $K$ the field of fractions of the ring of Witt vectors over $k$.

Question: If $X$ is an affine variety over $k$, do the rigid cohomology groups $H^i_\text{rig}(X/K)$ vanish for all $i > \dim X$?

This is true for smooth $X$, as in this case rigid cohomology agrees with Monsky-Washnitzer cohomology. So the question is about singular affine varieties. I'm mainly interested in $H^n_\text{rig}(X)$ for singular hypersurfaces $X$ in $\mathbb A^n$ over a finite field.

This question has a counterpart in characteristic 0:

If $Y$ is an affine variety over a field of characteristic $0$, then it is known that the $i$-th algebraic de Rham cohomology of $Y$ vanishes for $i > \dim Y$. Again, this is easy for smooth $Y$. In his article "On the de Rham cohomology of algebraic varieties", Hartshorne shows the vanishing for arbitrary affine $Y$ in two ways:

(1) For $i > \dim Y + 1$, one can use resolution of singularities and the Leray spectral sequence.

(2) For all $i$, there is an isomorphism (over $\mathbb C$) with the singular cohomology of the associated analytic space. Then apply Bloom-Herrera's result on the vanishing of singular cohomology of Stein spaces.

It seems that no algebraic proof is known for the case $i = \dim Y + 1$.

I wonder if there is some analogue of (2) in the rigid-analytic world. This could potentially give an affirmative answer to the question for rigid cohomology.

Another buzzword I want to throw in is "proper hypercoverings". Could they help?

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