Given $\mathrm{Gal}(E/ F)$ find $E$ and $F$

Question

There's a theorem that states that

for any finite group $G$ there exist a field $F$ with finite separable normal extension $E \supset F$ such that $G\cong \mathrm{Gal}(E/ F)$.

Is there an algorithm or a methods that allows to find such $E$ and $F$ with given $G$? I'm mostly interested in cases $G=V_4,D_4,C_4$.

Answer 1

$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$There is a result, called Artin' s Lemma that tells you the following.

Let $E$ be a field, and $G$ be a finite group of automorphisms of $E$. Let $F$ be the set of fixed points of $G$, that is, $$ F = \Set{ a \in E : a g = a \text{, for all $g \in G$}}. $$ Then $E/F$ is a Galois extension, with Galois group $G$.

Now let $G$ be a finite group. Without loss of generality, you may assume that $G \le S_{n}$ for some $n$. Let $$ E = \Q(x_{1}, \dots, x_{n}) $$ be the function field in $n$ variables. Then $G$ acts on $E$ as a group of automorphisms, by permuting the $x_{i}$ and we are done.

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Your special cases can be dealt with in an easier way.

For $V_{4}$, take $\Q(\sqrt{2}, \sqrt{3})/\Q$.

For $D_{4}$, take $\Q(\sqrt[4]{2}, i)/\Q$

For $C_{4}$, take $F(\sqrt[4]{2})/F$, where $F = \Q(i)$, or $\Q(\omega)/\Q$, where $\omega$ is a primitive fifth root of unity.

Answer 2

We can follow the usual construction for such a Galois extension $E/F$ from the proof of the Theorem: if $|G|=n$, then take $E=\mathbb{C}(x_1,\ldots ,x_n)$, and $k=\mathbb{C}(s_1,\ldots ,s_n)$, the subfield generated by the symmetric polynomials. Then $E/F$ is Galois, and by construction has $Gal(K/k)=S_n$. We have a canonical embedding of $G$ into $S_n$ by the group action of $G$ on itself by left multiplication. Let $H ⊂ S_n$ be the image of that embedding. By Galois correspondence, there exists an intermediate field $F$ so that $Gal(E/F) = H\cong G$.