The Quadratic Equation formula: Al-Khawarizmi (c 800 AD) did not have negative numbers, nor zeros, and also did not possess the needed algebraic notation. Therefore he had to devote 6 chapters of ...

No. The free profinite group $\widehat{\mathbb{Z}}$ on one generator is the direct product of the groups $\mathbb{Z}_p$, $p$ prime. Therefore each $\mathbb{Z}_p$ is a closed subgroup of $\widehat{\...

Here is one recent result in this direction, taken from I. Efrat and J. Minac, Galois groups and cohomological functors, Trans. of the AMS, http://www.ams.org/journals/tran/0000-000-00/S0002-9947-2016-...

Here is another example, which also answers the "followup question": Let $K$ be the field of Laurent series over $\mathbb{R}$. Its absolute Galois group is the infinite profinite dihedral group $\...

The fact that the Brauer group of a pseudo-algebraically closed field is trivial is noted in: J.Ax, The elementary theory of finite fields, Annals Math. 88 (1968), p. 269. A more detailed proof is ...

For $K=\mathbb{Q}$, this was proved byJ. Ax (Solving diophantine problems modulo every prime, Ann. Math. 85 (1967), 161-183). M. Jarden extended this to arbitrary Hilbertian fields (Algebraic ...

This space is studied in several papers by Haran, Jarden, and Pop, for instance: Projective group structures as absolute Galois structures with block approximation, Memoirs of AMS 189 (2007), 1--56 (...

I assume your field has characteristic $p>0$. Then the maximal pro-$l$ quotient of the absolute Galois group is torsion-free. Indeed, by results of E. Becker, Euklidische Korper und euklidische ...

In addition to the above very nice model-theoretic construction, it seems interesting to describe such fields Galois-theoretically. Here is one such construction: By a result of Ershov, the free ...

The field $k(t^\Gamma)$ is sometimes called "the field of generalized rational functions". It is covered in section 2.9 of I. Efrat, "Valuations, Orderings, and Milnor $K$-Theory", AMS, 2006.

Its torsion elements are only of order 2, and are all conjugate. This is a consequence of the fact that $\mathbb{Q}$ has exactly one ordering, using results of E. Becker (the pro-2 analog of the ...

One additional aspect may be worth mentioning: When trying to understand the structure of an absolute Galois group ${\rm Gal}(\overline{K}/K)$ of a field $K$, it is often helpful to keep in mind its ...

I understand question 2 as: Show that there is no Galois extension $F$ of $\mathbb{Q}$ with Galois group $C_4$ and such that $F$ contains $\mathbb{Q}(i)$. This is a special case of a general ...