I'm encountering a lot of problems when dealing with the root of unity sheaf $\mu_\infty := \mathrm{colim}_n\mu_n$.

Let $X$ be a smooth geometrically integral variety over a number field $k$. Although we have the canonical inclusion $\mu_\infty \subset \mathbb{G}_m$, the cohomology groups with coefficients in the latter sheaf are better understood. For example, in Galois cohomology for our given $k$ we know that $$H^1(k,\mathbb{G}_m) = 0,\,\,H^2(k,\mathbb{G}_m) = \mathrm{Br}(k), \,\, H^3(k, \mathbb{G}_m) = 0.$$

For etale cohomology on $X$, we have $$H^0(X,\mathbb{G}_m) = k[X]^*, \,\, H^1(X,\mathbb{G}_m) = \mathrm{Pic}(X), \,\, H^2(X,\mathbb{G}_m) = \mathrm{Br}(X).$$

Now consider the spectral sequence $$H^p(k,H^q(\bar{X},\mu_\infty)) \implies H^{p+q}(X,\mu_\infty).$$ One obtains a long exact sequence of low degree terms: $$0 \rightarrow H^1(k,H^0(\bar{X},\mu_\infty)) \rightarrow H^1(X,\mu_\infty) \rightarrow H^0(k,H^1(\bar{X},\mu_\infty)) \rightarrow H^2(k,H^0(\bar{X},\mu_\infty))$$

$$\rightarrow \mathrm{Ker}[H^2(X,\mu_\infty) \rightarrow H^0(k,H^2(\bar{X},\mu_\infty))] \rightarrow H^1(k,H^1(\bar{X},\mu_\infty)) \rightarrow H^3(k,H^0(\bar{X},\mu_\infty)).$$

**Question 2.** Does $H^2(\bar{X},\mu_\infty)$ have trivial Galois action? If so, is it true for all $H^i(\bar{X}, \mu_\infty)$?

This question came about because one of the papers I'm reading defined some term to be the kernel of $H^2(X,\mu_\infty) \rightarrow H^2(\bar{X},\mu_\infty)$, so I have a feeling it came from this spectral sequence.

There are many other questions I can think of but I would like to end off with this:

**Question 3.** Is there some connection between $H^1(X,\mu_\infty)$ and $\mathrm{Pic}(X)$?

**EDIT.** I've removed Question 1 due to some confusion.

Etale Cohomologyby Milne. This is my first year dealing with this topic, and I'm working without the completeness property of a variety, which is assumed in most of the book. Furthermore, I couldn't find any text dealing with the constant sheaf $\mu_\infty$, hence I asked these questions. I'm sorry if these three questions are too basic for the site, I can take them down. $\endgroup$3more comments