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Consider $x^{4}-2=(x+\sqrt[4]{2})(x-\sqrt[4]{2})(x+i\sqrt[4]{2})(x-i\sqrt[4]{2}) \in \mathbb{Q}[x]$. Let $K=\mathbb{Q}(\sqrt[4]{2},i)$ be the splitting field of $x^{4}-2$. Since $K$ is a splitting field and we are in characteristic 0, it follows that $K/\mathbb{Q}$ is Galois. Finally, $[K\colon\mathbb{Q}]=8$.

I want to compute the Galois group $K/\mathbb{Q}$. Since the extension is Galois, there are 8 elements in this group. It turns out that this group is isomorphic to the dihedral group of order 8 (I've seen examples of this but don't have a reference).

What steps would I take to reach the conclusion that this group is the dihedral group?

Specifically I know that each automorphism of the Galois group permutes the roots, but I don't see how to make the connection that these elements are the same as the dihedral group of order 8. I would appreciate a detailed analysis because I think this would allow me to apply these techniques to many other problems in Galois theory. Thanks.

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\sqrt[n]{a} produces $\sqrt[n]{a}$. –  Arturo Magidin Apr 20 '12 at 5:21
    
You can find in Lang's book on Algebra the study of extensions constructed by adjoining an $n$th root such as your $K/\mathbb Q$, and in particular the determination of their Galois groups, but if I recall correctly he only goes into detail for the case in which $n$ is odd, which is simpler. –  Mariano Suárez-Alvarez Apr 20 '12 at 6:07
    
\root n\of a also produces $\root n\of a$. –  Gerry Myerson Apr 20 '12 at 6:13
    
@GerryMyerson, that's really a plaintex-ism, which should probably be avoided in LaTeX sources. –  Mariano Suárez-Alvarez Apr 20 '12 at 6:22
    
@Mariano, I'm a PlainTeX kind of guy. Perhaps this comment thread isn't the place to discuss it, but I'd like to know why Plain constructs should be avoided here - they seem to be supported. –  Gerry Myerson Apr 20 '12 at 6:51
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2 Answers

up vote 5 down vote accepted

Since $x^4-2$ is irreducible over $\mathbb{Q}$ (Eisensein's Criterion at $p-2$, for example), the Galois group is transitive on the four roots. There is an automorphism that maps $\sqrt[4]{2}$ to $i\sqrt[4]{2}$; call it $\rho$. This map either maps $i$ to $i$, or $i$ to $-i$.

If it maps $i$ to $i$, then consider $\sigma$, complex conjugation. We have that $\sigma\rho$ maps $i$ to $-i$ and $\sqrt[4]{2}$ to $-i\sqrt[4]{2}$, whereas $\rho\sigma$ maps $i$ to $-i$ but $\sqrt[4]{2}$ to $i\sqrt[4]{2}$. So $G$ is not abelian.

If $\rho$ maps $i$ to $-i$, then $\sigma\rho$ maps $\sqrt[4]{2}$ to $-i\sqrt[4]{2}$ and $i$ to $i$. Taking $(\sigma\rho)^3$ we obtain a map that sends $i$ to $i$ and $\sqrt[4]{2}$ to $i\sqrt[4]{2}$, so we are back in the previous case. So either way, $G$ is not abelian.

That means that it is either dihedral, or the quaternion group of order $8$ (the only nonabelian groups of order $8$). But in the quaternion group, the only element of order $2$ is central (it is $-1$), whereas complex conjugation, which is of order $2$ in the Galois group, is not central, as we just saw. That means that the Galois group must be the dihedral group of order $8$. Alternatively, in the quaternion group there is a single element of order $2$; but $G$ has at least two: complex conjugation and $\rho^2$. So $G$ must be dihedral, not quaternion.

Explicitly, our map $\rho$ that sends $\sqrt[4]{2}$ to $i\sqrt[4]{2}$ and $i$ to $i$ is of order $4$; complex conjugation $\sigma$ is of order $2$; now note that $\sigma\rho=\rho^3\sigma$ (both map $i$ to $-i$, and $\sqrt[4]{2}$ to $-i\sqrt[4]{3}$). This gives you explicitly the dihedral structure: $G$ contains $\langle \sigma,\rho\mid \sigma^2=\rho^4=1,\ \sigma\rho=\rho^3\sigma\rangle$, which is of order $8$, so $G$ is this group, which is the dihedral group of order $8$.

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You don't say whether you know what the automorphisms are. If you do, you should be able to find automorphisms $f$ and $g$ with $f^4=1$, $g^2=1$, and $fg=gf^{-1}$, and that (together with knowing that $f$ and $g$ generate the group) is a presentation of the dihedral group.

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