Let $K_n = \Bbb Q(\sqrt[2^n]{2}, \zeta_{2^n})$ be a Galois extension of $\Bbb Q$ (where $ \zeta_{2^n}=e^{2\pi i / 2^n}$), and let $K$ be the compositum of all the $K_n$'s. It is a Galois extension of $\Bbb Q$. Notice that $m ≤ n \implies K_m \subset K_n$.

What does the Galois group of $K/\Bbb Q$ look like?

According to the proposition $1.1.$ of this document, $\text{Gal}(K/\Bbb Q)$ is the inverse limit of $\text{Gal}(L/\Bbb Q)$ where $L/\Bbb Q$ is a finite Galois extension such that $L \subseteq K$. In particular, we have to consider $L=K_n$, which has Galois group isomorphic to the "affine group" $\Bbb Z/2^n\Bbb Z \rtimes (\Bbb Z/2^n\Bbb Z)^{\times}$ (the holomorph of $\Bbb Z/2^n\Bbb Z$). What other subextensions should I consider? Moreover, I have some trouble as for understanding the inverse limit of all these Galois groups...

Some related questions are: (1), (2), (3). Here is a question with a similar infinite extension.

Any help would be highly appreciated. Thank you in advance!


First, it does suffice to just take the inverse limit of just your groups $\mathrm{Gal}(K_n/\mathbb Q)$, instead of including all finite Galois subextensions. This is a general fact, only requiring the $K_n$'s to be finite Galois extensions of the base field whose union is all of $K$. The proof is the same as in the notes you linked: an element of $\mathrm{Gal}(K/\mathbb Q)$ restricts to elements of $\mathrm{Gal}(K_n/\mathbb Q)$, giving a map $\mathrm{Gal}(K/\mathbb Q) \to \lim\limits_{\leftarrow n}\mathrm{Gal} (K_n/\mathbb Q)$. Then an element of $\mathrm{Gal}(K/\mathbb Q)$ is determined by its restrictions (giving injectivity), and compatible automorphisms of the $K_n$'s glue to an automorphism of $K$ (giving surjectivity).

Second, the inverse limit of your groups is the affine group over the $2$-adic integers, $\mathbb Z_2 \rtimes \mathbb Z_2^{\times}$. Proof. Regard the various affine groups as the groups of linear functions $mx + b$, where $m$ is invertible. (The group law is composition. Explicitly, for $K_n$, $mx + b$ is the automorphism sending $\zeta_{2^n}^x 2^{1/2^n}$ to $\zeta_{2^n}^{mx+b} 2^{1/2^n}$.) Notice that the restriction maps from $\mathrm{Gal}(K_{n+1}/\mathbb Q)$ to $\mathrm{Gal}(K_n/\mathbb Q)$ are the obvious quotient maps, given by modding $m$ and $b$ out by $2^n$.) Then $\mathbb Z_2 \rtimes \mathbb Z_2^{\times}$ maps (compatibly) to each $\mathbb Z/2^n \mathbb Z \rtimes \mathbb (Z/2^n \mathbb Z)^{\times}$ by taking $m$ and $b$ mod $2^n$, so it maps to the inverse limit. But this map is an isomorphism, because $\mathbb Z_2$ is (by definition) the inverse limit of $\mathbb Z/2^n \mathbb Z$ under the quotient maps, and $(\mathbb Z_2)^{\times}$ is the inverse limit of $(\mathbb Z/2^n \mathbb Z)^{\times}$. (The latter is because an element in either $\mathbb Z/2^n \mathbb Z$ or $\mathbb Z_2$ is invertible if and only if it is $1 \bmod 2$.)


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.