# First examples in Galois theory

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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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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