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I'm studying Field Theory and after studying theorems and problems about extensions, splitting fields, etc... I'm starting with the first theorems of the Galois Theory itself. In order to see if I understand such theorems, I'm trying to prove the first examples in Galois Theory such that the

Galois group of $x^3-2\in \mathbb Q[x]$ is the group of symmetries of the triangle.

I know that the roots of the equation of $x^3-2=0$ are $2^{1/3},2^{1/3}w,2^{1/3}w^{2}$, where $w$ is a root of the irreducible polynomial $x^2+x+1$ over $\mathbb Q(2^{1/3})$. Thus we write $x^3-2=(x-2^{1/3})(x-2^{1/3}w)(x-2^{1/3}w^{2})$, whence $E=\mathbb Q(2^{1/3},w)$. It follows that $[E:\mathbb Q]=6$. Since E is a splitting field and therefore normal we have also $G(E/\mathbb Q)=6$, then we have six automorphisms of $E$.

I'm stuck here, I can't go further, I need help please.

Thanks a lot.

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4 Answers 4

up vote 6 down vote accepted

Since you know that $|G(E/\mathbb Q)|=6$ and since the Galois group permutes the roots, of which there are three, it follows that the Galois group is some group with 6 elements that can be identified with a subgroup of $S_3$. That narrows it down to precisely $S_3$ (up to isomorphism).

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The easiest way to solve this particular problem is to note that $G(E/\mathbb{Q})$ is nonabelian and has order $6$, from which the answer follows immediately. Below is a constructive method to determine this, which should hopefully also give you insight into how to approach future problems of this nature.

Let $\sigma\in G(E/\mathbb{Q})$. Since $E=\mathbb{Q}(2^{1/3},\omega)$, it suffices to define $\sigma$'s action on $E$ by defining its action on $\omega$ and $2^{1/3}$.

First, we consider $\sigma(2^{1/3})$. $\sigma$ must send conjugates to conjugates - that is, if $x\in E$, the minimal polynomial of $x$ must be the same as the minimal polynomial of $\sigma(x)$. Therfore, we must have $\sigma(2^{1/3})=2^{1/3}$, $\omega 2^{1/3}$, or $\omega^2 2^{1/3}$. Similarly, $\sigma(\omega)$ must go to $\omega$ or $\omega^2$. (Note that $\sigma(\omega)$ can't go to $\omega^3=1$, since this would not fix $\mathbb{Q}$.)

Intuitively we see it's best to define $$\rho:\left\{\begin{array}{l}2^{1/3}\mapsto \omega 2^{1/3}\\\omega\rightarrow\omega\end{array}\right. \text{ and }\tau:\left\{\begin{array}{l}2^{1/3}\mapsto 2^{1/3}\\\omega\rightarrow\omega^2\end{array}\right.$$ We can easily verify that $\rho$ and $\tau$ are indeed automorphisms of $E$ fixing $\mathbb{Q}$, and therefore that $\langle \sigma,\tau \rangle\leqslant G(E/\mathbb{Q})$. Indeed, by design we can write any admissible $\sigma\in G(E/\mathbb{Q})$ in terms of $\sigma$ and $\tau$, so we expect $\langle \sigma,\tau \rangle = G(E/\mathbb{Q})$. We confirm by matching the presentation of $\langle \sigma,\tau \rangle$ to $$D_3=\langle r,s|r^3,s^2,r^s=r^{-1}\rangle$$ that $\langle \sigma,\tau \rangle = G(E/\mathbb{Q})=D_3$.

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A possible next step is to determine what those 6 automorphisms are. $w$ can only go to a root of $x^2+x+1$ (two choices), and $2^{1/3}$ can only go to a root of $x^3-2$ (three choices). This gives at most $2 \cdot 3 = 6$ automorphisms, and since we know there are 6 automorphisms, these must be all of them.

Now verify that these automorphisms actually generate a group isomorphic to $S_3$.

After that, you might try to determine the subgroups of $S_3$, and the corresponding subfields of $E$ according to the Fundamental Theorem of Galois Theory.

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look at the automorphism

$\sigma(2^\frac{1}{3}) = 2^\frac{1}{3}w$ and $\sigma(w) = w$ (as $2^\frac{1}{3}$ can go to $2^{1/3},2^{1/3}w,2^{1/3}w^{2}$)

$\tau(2^\frac{1}{3}) = 2^\frac{1}{3}$ and $\tau(w) = w^2$ (as $w $ ca go to $w $ and $w^2$)

then <$\sigma,\tau$> =$S_3$

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