I'm having a bit of trouble understanding an example from the introduction to galois theory section from Gallian's text.

We have that $\omega = -1/2 + i\sqrt 3/2$ satisfies the equations $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$. We now consider the extension $Q(\omega,2^{1/3})$ of $Q$.

I understand that any automorphism of a field containing Q must fix Q, and since the basis of the extension field will involve powers and products of $\omega$ and $2^{1/3}$, we just need to find out where the automorphisms send those elements.

What I don't understand is how we determine that there are six possible automorphisms and where they send these elements. The automorphisms he lists in the text send

$\omega \to \omega$ or $\omega^2$ and $2^{1/3} \to 2^{1/3}$ or $\omega(2^{1/3})$ or $\omega^2(2^{1/3})$.

I'm just having a bit of trouble seeing how he determined where the automorphism sends the elements, I suspect it has something to do with the basis of the extension? But I'm not sure.

In a previous example when we had the extension of $Q(\sqrt 2)$ over $Q$, the automorphisms could send $\sqrt 2$ to $\pm \sqrt 2$ since

$2 = \phi(2) = \phi(\sqrt 2 \sqrt 2) = (\phi(\sqrt 2))^2$

In this case the basis of the extension is $\{1,\sqrt 2\}$, so we're not just sending the basis element $\sqrt 2$ to another basis element, rather its conjugate.

I'm guessing there's a fundamental misunderstanding I'm having about determining the automorphisms.


  • 1
    $\begingroup$ Any automorphism sends an element to its conjugates. That's how author is determining those automorphism. Also if you are familiar with Galois extensions and splitting fields , in this case you can determine the order of group of automorphisms (Galois group) without explicitly calculating the maps. $\endgroup$ – iron feliks Apr 21 '16 at 18:09
  • $\begingroup$ As @skmehta says, $\omega$ must go to a root of $X^2+X+1$, and your chosen cube root of $2$ (you’ve called it $2^{1/3}$) must go a root of $X^3-2$. There’s your six members of the Galois group. $\endgroup$ – Lubin Apr 21 '16 at 19:45

The extension $\Bbb Q(\sqrt[3]{2},\omega)$ is the splitting field of the irreducible polynomial $x^3 - 2 \in \Bbb Q[x]$. It is not hard to see that:

$x^3 - 2 = (x - \sqrt[3]{2})(x - \sqrt[3]{2}\omega)(x - \sqrt[3]{2}\omega^2)$

that is, each of: $\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2$ is a cube root of $2$.

Part of the behavior of the Galois group of $x^3 - 2$ has to do with the "threeness of three", an odd degree rational polynomial has either one, or three real roots. This is because complex-conjugation (the algebraic closure of $\Bbb Q$ is a subfield of $\Bbb C$) is a field automorphism of $\Bbb C$ that fixes the reals (and thus the rationals).

In this case, we have the partial factorization over $\Bbb Q(\sqrt[3]{2})$:

$x^3 - 2 = (x - \sqrt[3]{2})(x^2 + \sqrt[3]{2}x + \sqrt[3]{4})$

and we have "two types" of automorphisms of $\Bbb Q(\sqrt[3]{2},\omega)/\Bbb Q$: ones that send all three roots of $x^3 - 2$ "in a circle" (cycle), and ones that swap just two of them (with a quadratic only the second type is possible).

Since we have 3 roots, we have "some" subgroup of $S_3$, and since this group has a 3-cycle AND a two-cycle (we know it has a two-cycle, because the restriction of complex conjugation to $\Bbb Q(\sqrt[3]{2},\omega)$ is obviously an automorphism of $\Bbb Q(\sqrt[3]{2},\omega)/\Bbb Q$ of order 2), it must be $S_3$. From here it's just a matter of "back-figuring" what each permutation of the roots must do to the generators. That is:

$\text{Gal}(\Bbb Q(\sqrt[3]{2}/\Bbb Q) = \langle \sigma, \tau\rangle$ where:

$\sigma: \sqrt[3]{2}\mapsto \sqrt[3]{2}\omega \mapsto \sqrt[3]{2}\omega^2 \mapsto \sqrt[3]{2}$

$\tau: \sqrt[2]{2}\omega \mapsto \sqrt[3]{2}\omega^2 \mapsto \sqrt[3]{2}\omega$.

The roots of $x^3 - 2$, when plotted on the complex plane, form the vertices of an equilateral triangle, making it appear less mysterious that their symmetry group is the symmetry group of said triangle.


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