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Write an example of a Galois extension of fields that has as a Galois group $(\mathbb{Z}/2\mathbb{Z})^{3}$


I'm not very familiar with Galois theory, so I don't know of a general procedure to determine a Galois extension as having a specific Galois group. I was thinking of perhaps $\mathbb{Q(\sqrt {2},\sqrt{3},\sqrt{5})}$, but this seems unlikely and kind of complicated.

What would be a simpler example for this? Thanks!

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    $\begingroup$ Your example is good, see this question $\endgroup$
    – lulu
    Mar 28, 2016 at 0:17

2 Answers 2

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Given a subgroup $G \subseteq S_n$ (and every finite group occurs as a subgroup like this), it is not too hard to find a Galois extension $L/K$ with $L = \mathbb{Q}(x_1,...,x_n)$ and $\mathrm{Gal}(L/K) \cong G,$ by letting $K$ be the subfield fixed by $G$ under the action of permuting the variables $x_1,...,x_n.$

There is no general procedure to determine a Galois extension of $\mathbb{Q}$ with a given Galois group. For abelian extensions it's easier, though - every finite abelian group occurs as a subgroup (and therefore a quotient group) of some $(\mathbb{Z}/n \mathbb{Z})^{\times}$, so you can find a subfield of $\mathbb{Q}(\zeta_n)$ that will work.

One example of an extension of $\mathbb{Q}$ with that Galois group is $\mathbb{Q}(\zeta_{24})$.

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The other answer is much more complete but let me just add this nice class of examples. You probably know that if $L=\mathbb Q(\sqrt n)$ for a squarefree integer $n$, then Gal$(K/\mathbb Q)\cong \mathbb Z/2\mathbb Z$. Now try to show that for distinct primes $p_1,p_2\cdots p_n$,

$$\text{Gal}(\mathbb Q(\sqrt p_1,\sqrt p_2\cdots ,\sqrt p_n)/\mathbb Q)\cong (\mathbb Z/2\mathbb Z)^n. $$

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