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Prove or disprove:

Claim: there exists a field $K$ and an integer $n$ such that $\pi_1^{et}((GL_n)_K)$ is isomorphic to the profinite completion of the abstract group $GL_n(K)$?

Note that then $G_K \cong \widehat{GL_n(K)}/\widehat{\mathbb{Z}}$. In particular, $G_K$ must have a (topological) generating set of size the cardinality of $K$.

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    $\begingroup$ The conclusion that $G_K = \widehat{GL_n}(K)/\widehat{Z}$ (or rather $\widehat{Z}(1)$?) is only true if $K$ has characteristic zero. What is the motivation for this question? $\endgroup$
    – Piotr Achinger
    Commented Dec 19, 2018 at 9:30
  • $\begingroup$ What is $\widehat{GL_n(K)}$ anyway? If $K$ has characteristic zero, then any unipotent element of $GL_n(K)$ is divisible in the sense that it is in the image of a group homomorphism $\mathbf{Q}\to GL_n(K)$, and therefore such elements have to lie in the kernel of $GL_n(K)\to \widehat{GL_n(K)}$. Thus the subgroup $G\subseteq GL_n(K)$ generated by unipotent elements, which is likely to be all of $SL_n(K)$ (didn't check), lies in the kernel. So it seems that $\widehat{GL_n(K)} = \widehat{ K^*}$. $\endgroup$
    – Piotr Achinger
    Commented Dec 19, 2018 at 12:56
  • $\begingroup$ The fundamental group also doesn’t depend on $n$, I think. If $K$ is of char 0, the determinant map induces isomorphism on fundamental groups(because it does so over $\bar{K}$) and for $GL_1$ it is just the semidirect product $G_K\ltimes \hat{\mathbb{Z}}(1)$. If we want this group to be isomorphic to the completion of $K^{\times}$, it should be abelian so the action which is used to form the semidirect product has to be trivial. That is, $K$ contains all the roots of unity and its absolute Galois group is abelian. $\endgroup$
    – SashaP
    Commented Dec 19, 2018 at 14:38
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    $\begingroup$ In this case, by Kummer theory the absolute Galois group is precisely the profinite completion of $K^{\times}$. So a field $K$ satisfies your condition iff it contains all the roots of unity, has abelian absolute Galois group and its Galois group(which is isomorphic to completed group of units) has the following telescopic property: $G_K\cong G_K\times \hat{\mathbb{Z}}$. An example of such is $K=\mathbb{C}((t_1, t_2,\dots ))$(the direct limit of the fields of Laurent series over an infinitely increasing set of variables) whose Galois group is the product of infinitely many copies of &\hat{Z}$. $\endgroup$
    – SashaP
    Commented Dec 19, 2018 at 14:46
  • $\begingroup$ @SashaP why is the Galois group of $\mathbb{C}((t_1, t_2, \ldots))$ the direct product as you say? You cannot use Kunneth here, but maybe you meant something different? $\endgroup$
    – Piotr Achinger
    Commented Dec 19, 2018 at 14:58

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