Galois group of $x^3 - 2 $ over $\mathbb Q$

Solution 1:

A brute-force way to see it is easy enough. The roots of $X^3 - 2$ are $\sqrt[3]{2},\; \omega\sqrt[3]{2},\; \omega^2\sqrt[3]{2}$, so the splitting field of your polynomial is $K = \mathbb{Q}(\sqrt[3]{2},\omega)$. You are asking why it's legitimate to send $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$. Well, let's try it. Let $\sigma: K \to K$ be a map that sends $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$. How are we going to extend this to an element of the Galois group? We need $\sigma$ to be an automorphism that fixes $\mathbb{Q}$.

Our field $K$ is a $\mathbb{Q}$-vector space generated by six basis elements: $1,\; \sqrt[3]{2},\; \sqrt[3]{4},\; \omega,\; \omega\sqrt[3]{2},\; \omega\sqrt[3]{4}$. If we can define how $\sigma$ acts on those, we can extend linearly to the whole space: that is, we can extend $\sigma$ such that $\sigma(a+b) = \sigma(a) + \sigma(b)$ for all $a$ and $b$. We also know that, if we define $\sigma$ sensibly, we should get that $\sigma(ab) = \sigma(a) \sigma(b)$. By "sensibly" here, I mean that $\sigma$ should send each element of $K$ to one of its conjugates - that is, it should send $x$ to any $y$ that satisfies the same minimal polynomial as $x$. We also know that it's enough to make sure $\sigma$ is multiplicative on the basis, and in fact we only need to define it on $\sqrt[3]{2}$ and $\omega$, since the rest will then follow automatically by multiplicativity.

So we have a few options from what we've deduced so far. $\sigma$ can send $\sqrt[3]{2}$ to any of $\sqrt[3]{2},\; \omega\sqrt[3]{2},\; \omega^2\sqrt[3]{2}$, and it can send $\omega$ to either of $\omega$ and $\omega^2$. So let's say $\sigma(\sqrt[3]{2}) = \omega\sqrt[3]{2}$ (as we wanted) and $\sigma(\omega) = \omega$ (no good reason for this choice, but we had to make one). Where do all the other basis elements go? Can you convince yourself now that $\sigma$ is an element of the Galois group?

(What would have happened if we'd chosen $\sigma(\omega) = \omega^2$?)

Solution 2:

What you're probably missing is that, although the real cube root of 2 stands out, all three complex cube roots of 2 are algebraically the same, that is, they cannot be told apart, and so can be permuted at will. Hence $S_3$.

Contrast this with the seemingly analogous situation of, say, the 6-th roots of unity. Here the roots are not the same algebraically, because $x^6-1$ is not irreducible.