1
$\begingroup$

I am trying to understand infinite galois theory by self-made examples. First example I am struggling with is the field of all constructible (by compass and straightedge) complex numbers - let's call it $\mathbb{M}$. Well known fact is, that every such constructible number has degree $2^k$ over $\mathbb{Q}$ for some $k\in \mathbb{N}$. In fact there is, for every $\alpha\in\mathbb{M}$ a chain of fields $\mathbb{Q} = K_0 \subset K_1 \subset \dots K_m$ with $\alpha\in K_m$ and $[K_n:K_{n-1}] = 2\ \forall n=1, \dots, m$.

I do wonder now, what the Galois group of $\mathbb{M}/\mathbb{Q}$ looks like. I am aware of the construction of the Krull topology on the projective limit of all finite Galois subextensions, but precisely in order to understand this construction, I am making this example.

So here's my first question, from which I hope I can proceed to understand: can someone name a simple example of an element of the Galois group? Is for example a map, that maps each instance of $\sqrt2$ to $-\sqrt2$ a $\mathbb{Q}$-automorphism of $\mathbb{M}$ and therefore an element of the group? If so, how to show that? Part of the problem here seems to be, that I don't even know, how to write the elements of $\mathbb{M}$.

Now if this is the case and my map is an element of $Gal(\mathbb{M}/\mathbb{Q})$, how does it act on intermediate galois fields like $\mathbb{Q}(\sqrt6)$ or $\mathbb{Q}(i, \sqrt{\sqrt{2}+1})$? Are these expressions "opaque" for this automorphism or is it rather as I (naively) expected, that each number that "contains" $\sqrt2$ is affected by this element?

What I want to get at is this: suppose we fix some intermediate, finite Galois extension fields as the first components of the infinite Cartesian product of which the projective limit is a subset, say $\mathbb{Q}(\sqrt2)$ as the first, then $\mathbb{Q}(\sqrt3)$, $\mathbb{Q}(\sqrt6)$ and $\mathbb{Q}(i, \sqrt{\sqrt{2}+1})$. The isomorphism between this product and $Gal(\mathbb{M}/\mathbb{Q})$ maps an element $\varphi \in Gal(\mathbb{M}/\mathbb{Q})$ to the vector of the restrictions of $\varphi$ to all finite Galois extension fields. So now, what is the image of a possible $\varphi: (\sqrt2 \mapsto -\sqrt2)$ under this isomorphism in the first components?

I am hoping that in the long run, this example will enable me, to understand, which fields are open and which closed in the Krull topology.

$\endgroup$
  • $\begingroup$ You have to consult maybe (if you don't know it) the excelent book "L'Aritmetique des corps" of Paulo Ribenboim, Chapter VII. Almost sure there is an edition in english. Your (not easy) problem is motivating. $\endgroup$ – Piquito Apr 28 '15 at 18:39
  • $\begingroup$ have you looked at the construction of projective limits ? because it should answer most of your questions. $\endgroup$ – mercio Apr 28 '15 at 20:28
  • $\begingroup$ I rather think that there will be several automorphism $\phi$ of $M$ with the property $\phi(\sqrt2)=-\sqrt2$. A tool I would use here is the following. If $K$ is a perfect field and $\overline{K}$ an algebraic closure. If $\alpha\in \overline{K}\setminus K$ shares its minimal polynomial with $\beta\in\overline{K}$, then the known isomorphism $f:K[\alpha]\to K[\beta]$ determined by $f(\alpha)=\beta$ extends (possibly non-uniquely) to an automorphism of $\overline{K}$. The proof is a usual Zorn's lemma business. $\endgroup$ – Jyrki Lahtonen Apr 29 '15 at 6:45
  • $\begingroup$ Hmm. A more general version may be needed? If $L_1, L_2$ are two isomorphic intermediate fields, $\sigma:L_1,L_2$ an isomorphism, $m(x)\in L_1(x)$ the minimal polynomial of $\alpha\in\overline{K}\setminus L_1$, and $\beta\notin L_2$ a zero of $m^\sigma(x)\in L_2[x]$, then $\sigma$ can be extended to an isomorphism of $\overline{K}$ such that $\sigma(\alpha)=\beta$. $\endgroup$ – Jyrki Lahtonen Apr 29 '15 at 6:48
  • $\begingroup$ So in your setting any automorphism of any subfield of $M$ extends to an automorphism of $\overline{\Bbb{Q}}$ (in possibly infinitely many ways). Restricting those extensions to $M$ then gives you things to play with. It seems to me that $M$ is Galois over $\Bbb{Q}$, i.e. stable under all those restrictions of automorphisms of $\overline{\Bbb{Q}}$. $\endgroup$ – Jyrki Lahtonen Apr 29 '15 at 6:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.