Let $X$ be a smooth and irreducible variety over a field.
Does the Zariski cohomology $H^n(X,\mathbb{Z}[\mathbb{G}_m])$ vanish for $n>1$?
Here $\mathbb{Z}[-]\colon Set\to Ab$ is (the Zariski sheafification of) the free abelian group functor applied sectionwise to the sheaf $\mathbb{G}_m$.
This guess comes from the analogous situation ''without'' the free abelian group which is explained here: We have $$ H^n(X,\mathbb{G}_m) = 0 \quad \mbox{for $n>1$} $$ by the following argument. The exact sequence $$ 1\to \mathbb{G}_m \to K_X^\times \to \mathrm{CaDiv} \to 1 $$ of Zariski sheaves (defining Cartier divisors $\mathrm{CaDiv}$) is a resolution by flasque sheaves: The constant sheaf $K_X^\times$ is flasque and since Cartier divisors are Weil divisors $\mathrm{CaDiv}=\mathrm{WDiv}$ on the regular $X$, the sheaf $\mathrm{CaDiv}$ is flasque as well.
Now also $\mathbb{Z}[K_X^\times]$ and $\mathbb{Z}[\mathrm{WDiv}]$ are flasque Zariski sheaves but the sequence $$ 1\to \mathbb{Z}[\mathbb{G}_m] \hookrightarrow \mathbb{Z}[K_X^\times] \twoheadrightarrow \mathbb{Z}[\mathrm{WDiv}] $$ may not be an exact sequence of Zariski sheaves, I guess: One only gets a stalkwise surjection $$ \mathbb{Z}[K_X^\times]/\mathbb{Z}[\mathbb{G}_m] \twoheadrightarrow \mathbb{Z}[\mathrm{WDiv}] $$ (Note that although $\mathbb{Z}[-]$ preserves colimits/quotients of sets, this may not be true for quotients in abelian groups) and I don't have an idea on how to control the kernel. Maybe, there is an easier way to see that $$ \mathbb{Z}[K_X^\times]/\mathbb{Z}[\mathbb{G}_m] $$ is a flasque Zariski-sheaf. Does anybody have an idea? Thank you!