By an *arithmetic scheme* I mean a finite type flat regular integral scheme over $\mathrm{Spec} \, \mathbb{Z}$.

Let $X$ be an arithmetic scheme. Then is $H_{et}^2(X,\mathbb{Z}/n\mathbb{Z})$ finite for all $n \in \mathbb{N}$?

Remarks: Let $j:U \to X$ be the open subset given by removing the fibres of $X \to \mathrm{Spec} \, \mathbb{Z}$ lying above those primes dividing $n$.

- I can easily show that $H_{et}^1(X,\mathbb{Z}/n\mathbb{Z})$ is finite. For example, it follows from the Leray spectral sequence that $H_{et}^1(X,\mathbb{Z}/n\mathbb{Z}) \to H_{et}^1(U,\mathbb{Z}/n\mathbb{Z})$ is injective, and the latter group is finite by a special case of [1, Proposition II.7.1].
- If $n$ invertible on $X$, then the answer is
*yes*and this is again a special case of [1, Proposition II.7.1]. So I'm interested in the case where $n$ is not invertible on $X$. (In particular, this shows that $H_{et}^2(U,\mathbb{Z}/n\mathbb{Z})$ is finite). - If $\dim X = 1$ then the answer if
*yes*by [1,Theorem II.3.1]. - Another application of the Leray spectral sequence argument shows that is suffices to show that $H^0(X, R^1j_* \mathbb{Z}/n\mathbb{Z})$ is finite, but I don't see why this should be the case.
- The analogous question concerning finiteness of $H_{et}^i(X,\mathbb{Z}/n\mathbb{Z})$ is also interesting, but for my application I only need $i = 2$.

References:

[1] Milne - Arithmetic duality theorems.