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(I asked this question a couple of days back on Stackexchange but with no success, it seems elementary, but I am struggling to go about attempting it.)

Let $X$ be a smooth geometrically integral variety over a number field $k$. We denote by $\bar{k}[X]^*$ the group of invertible functions on $\bar{X}$, and let $$G = \varinjlim_{n \in \mathbb{Z}_{>0}}I_n$$ where $I_n=\bar{k}[X]^*$ for all $n$ and for $m$ such that $m|n$, we have $$I_m \rightarrow I_n:\bar{k}[X]^* \rightarrow \bar{k}[X]^*:x \mapsto x^{n/m}.$$ Question 1. How do we show that this direct limit is $\bar{k}[X]^* \otimes_\mathbb{Z} \mathbb{Q}$? Most examples of computing direct limits are related to multiplicative maps between $\mathbb{Z}$, so I'm struggling to find any similarities in methods.

Furthermore, we have $$H^i(k,\bar{k}[X]^*\otimes \mathbb{Q}) \cong H^i(k,\bar{k}[X]^*) \otimes \mathbb{Q},$$ where $H^i(k,-)$ denotes the Galois cohomology functor.

Question 2. How do we compute $H^0(k,\bar{k}[X]^*)$, i.e., the invertible functions fixed by Galois action?

Question 3. I believe that $H^i(k,\bar{k}[X]^*)$ is torsion for $i \geq 1$, is there any direct way to show this?

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    $\begingroup$ Are you aware that $k[X]^*$ ($k$ doesn't need to be algebraically closed) has a very nice group structure? In fact $k[X]^*/k^*$ is free abelian group of finite rank. $\endgroup$
    – user127776
    Commented Jul 18, 2021 at 7:46
  • $\begingroup$ @user127776 most of what I encountered has the proper condition imposed on $X$, so $\bar{k}[X]^*$ is simply $\bar{k}^*$. I've never thought about the structure of $k[X]^*$ itself, could you explain it further? Also, the quotient $k[X]^*/k^*$ is finitely generated, I didn't know about it being free. $\endgroup$
    – oleout
    Commented Jul 18, 2021 at 7:51
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    $\begingroup$ Your last question is asking about $H_{et}^i(k,\mathbb{G}_m)$. It is torsion for $i\geq1$. Even if you replace $k$ by $X$ it is true for $i>1$. In the case of fields it can be shown it is also torsion for $i=1$. The easiest way that I can explain this is by using motivic cohomology and the fact $\mathbb{G}_m[-1]$ is the weight one etale motivic complex which rationally agrees with the motivic cohomology. But I am sure there is an elementary proof too, see page 88 here: jmilne.org/math/CourseNotes/LEC.pdf $\endgroup$
    – user127776
    Commented Jul 18, 2021 at 8:39
  • $\begingroup$ Alright, I will have a look at it, thanks. $\endgroup$
    – oleout
    Commented Jul 18, 2021 at 9:09

1 Answer 1

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As for the first question:

It is an elementary exercise that $\mathbb{Q}$ is the colimit of the diagram of copies of $\mathbb{Z}$ indexed by the positive integers with the divisibility relation. The transition maps are given by multiplying with the corresponding fraction, and a number $z$ in the $n$th copy represents $z/n$.

Since the tensor product is cocontinuous in both variables, it follows that for every abelian group $A$ the tensor product $A \otimes \mathbb{Q}$ is the colimit of the diagram of copies of $A \otimes \mathbb{Z} \cong A$ with the evident transition maps.

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  • $\begingroup$ So just to clarify what you've described: let $\mathbb{Z}_{(n)}$ denote the $n$-th copy of $\mathbb{Z}$. The elements of $\mathbb{Z}_{(n)}$ are of the form $i/n$ for all $i \in \mathbb{Z}$. We define the map $\mathbb{Z}_{(n)} \rightarrow \mathbb{Z}_{(mn)}$ by sending $x \in \mathbb{Z}_{(n)}$ to $x \cdot n/mn = x \cdot 1/m = x/m$. If this is wrong please let me know, otherwise, thank you for your input, I can accept this as the answer as this is the main question, as described in the title. $\endgroup$
    – oleout
    Commented Jul 19, 2021 at 10:54

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