I'm looking at the Galois group $\mathrm{G}(\mathbf{K}/\mathbb{Q})$ where $\mathbf{K}$ is the splitting field of $x^3-2$ over $\mathbb{Q}$. Of course, $\mathbf{K} = \mathbb{Q}(\alpha_1, i\sqrt{3})$, letting $\alpha_1$ be the real root of the polynomial (i.e. $\sqrt[3]{2}$) and letting $\alpha_2, \alpha_3$ be the two imaginary roots. Then we can define any of the automorphisms of $\mathrm{G}(\mathbf{K}/\mathbb{Q})$ by its action on the $\alpha_i$ and $i\sqrt{3}$ . For instance, define the automorphism: $$\Psi_{1,2} : \alpha_1 \to \alpha_2, \; i\sqrt{3} \; fixed.$$

And similarly we can obtain all the other elements of $\mathrm{G}(\mathbf{K}/\mathbb{Q})$. My question is then, since we only defined $\Psi_{1,2}$ in terms of its action on $\alpha_1$, how do we determine the action of this $\Psi_{1,2}$ on $\alpha_2$ and $\alpha_3$ without actually computing numerically? That is, without using the fact that we know the numerical values of the $\alpha_i$? Can we?

Some preliminaries we know: $\Psi_{1,2}$ maps roots of polynomials onto other roots of the same polynomial, so I really only need to figure out if $\Psi_{1,2}(\alpha_2)$ is $\alpha_1$ or if it is $\alpha_3$.

I've tried using the properties of $\mathbf{K}$ as a vector space over $\mathbb{Q}$:

$$\Psi_{1,2}(\alpha_2) = \Psi_{1,2}(a_0 + a_1\alpha_1 + a_2\alpha_1^2) = a_0 + a_1\alpha_2 + a_2\alpha_2^2,$$ where $a_i \in \mathbb{Q}.$

I subsequently got into an arithmetic/algebraic mess involving systems of equations, so I'm wondering if I'm on the right track or should try a different approach. Thanks.


It's easier to see what happens if you choose a different (but equivalent) basis. While it's true that $\alpha_1$ (the real root of $x^3 - 2$) and $i \sqrt{3}$ generate $K$ over $\mathbb{Q}$, it's a little artificial to see how the automorphisms work in terms of this basis.

Instead, use $\omega = \frac{-1 + i \sqrt{3}}{2}$. $\omega$ is one of the two non-real cube roots of unity (it's a root of $x^2 + x + 1 = 0$, which is $(x^3 - 1)(x-1)^{-1})$, and $\omega^2 = \frac{-1 - i \sqrt{3}}{2}$. Clearly, $\omega$ and $i\sqrt{3}$ generate the same extensions, since they are inter-definable in terms of $\mathbb{Q}$. Usually, we like studying quadratic extensions by taking the generator of the form $\sqrt{D}$, but in this case, the algebra works more nicely with $\omega$.

To see why, notice that the three roots of $x^3 -2$ are $\alpha_1, \omega \alpha_1 = \alpha_2$, and $\omega^2 \alpha_2 = \alpha_3$, since for example $(\omega \alpha_1)^3 = \omega^3 \alpha_1^2 = 2$.

Thus, the automorphism $\Psi_{1,2}$ of $K$ which maps $\alpha_1$ to $\alpha_2$ is really mapping $\alpha_1$ to $\omega \alpha_1$. Since it fixes $i \sqrt{3}$, it fixes $\omega$ as well (since it fixes $\mathbb{Q}$), so $\Psi_{1,2}(\alpha_2) = \Psi_{1,2}(\omega)\Psi_{1,2}(\alpha_1) = \omega^2 \alpha_1$. Similarly, $\Psi_{1,2}(\alpha_3) = \Psi_{1,2}(\omega^2 \alpha_1) = \omega^3 \alpha_1 = \alpha_1$. Thus $\Psi_{1,2}$ acts on the roots of $x^3 - 2$ by the cyclic permutation $(1 2 3)$. Similarly, the automorphism which maps $\alpha_1$ to $\alpha_3$ and fixes $\omega$ is the cyclic permutation $(132)$.

Now, the map which takes $i \sqrt{3}$ to $- i \sqrt{3}$ and fixes $\alpha_1$ interchanges $\omega$ and $\omega^2$ (as you can see from their explicit formulas). Thus, it maps $\alpha_2 = \omega \alpha_1$ to $\omega^2 \alpha_1 = \alpha_3$ and similarly it maps $\alpha_3$ to $\alpha_2$. Thus it is the transposition $(23)$.

Combining this information, we see that our Galois group can be identified with the subgroup of $S_3$ generated by $(123), (132)$, and $(23)$. This must be $S_3$ (which we also know because we know the extension has degree $6$).

  • 1
    $\begingroup$ I like this a lot, ’cause it’s exactly what I would have said (except that you’ve probably said it better than me). $\endgroup$ – Lubin Sep 4 '15 at 17:32
  • $\begingroup$ Wow, that means a lot. Thanks! :) $\endgroup$ – Dorebell Sep 4 '15 at 19:04
  • $\begingroup$ Wow, beautiful - thanks so much. Is there a way to kind of tell right off the bat that I should have used $\omega$ as a basis element instead of $i\sqrt{3}$? $\endgroup$ – Chris Sep 4 '15 at 21:18
  • $\begingroup$ Well it's a root of unity, and any time your field extension involves roots of unity, it's a good idea to think in terms of them. Extensions generated by roots of unity (called cyclotomic extensions) are among the best-understood and simplest there are. Also, when you're discussing the splitting field of a polynomial, it's a good idea to pay attention to any symmetries the polynomial has. For a polynomial of the form $x^n-a$, any two roots differ by multiplication by a root of unity, and this is a wonderful symmetry to use. $\endgroup$ – Dorebell Sep 4 '15 at 21:27

You can, but I am not sure if that is too important to do. What might be more relevant is the general group structure of ${\rm Gal}(K/\Bbb Q)$ -- that is, how the automorphisms act on each other, and not what they do elementwise. In general, suppose that $f=X^p-a$ is irreducible over $\Bbb Q[X]$. A splitting field for $f$ over the rationals is $E=\Bbb Q(\xi,r)$ where $r^p=a$. A general useful but easy to prove fact is that

Suppose that we have a Galois extension $E/F$ with subextensions $L,K$ such that $LK=E$ and $L\cap K=F$, $L/F$ a normal extension. Then $G={\rm Gal}(E/F)$ is a semidirect product of $N={\rm Gal}(E/L)$ with ${\rm Gal}(E/K)$, i.e. $G=NH$, $N\lhd G$, $H\cap H=1$.

Assuming that $f$ is irreducible amounts to the fact $f$ that has no roots over $\Bbb Q$. With the above in mind, consider the subextensions $\Bbb Q(r\xi)=K,L=\Bbb Q(\xi)$. These generate $E$ and are of degree $p,p-1$ respectively so they must interesct trivially when $p$ is odd. You can show that $N=\langle \tau\rangle$ and $H=\langle \sigma\rangle$ with $\tau(\xi)=\xi$, $\tau(r)=r\xi$ and $\sigma(\xi)=\xi^i,\sigma(r)=r \xi^{-i}$ where $i$ is a generator of the cyclic group $\Bbb Z_p^\times$, i.e. its powers generate the units of $\Bbb Z_p$.

A calculation (now elementwise!) shows that $\sigma$ has order $p-1$, $\tau$ has order $p$ and $\sigma\tau\sigma^{-1}=\tau^i$. This determines ${\rm Gal}(E/\Bbb Q)$ completely, and you can show it is isomorphic to the holomorph of $\Bbb Z_p$, namely the semidirect product $\Bbb Z_p\rtimes \Bbb Z_p^\times$ where $\Bbb Z_p^\times ={\rm Aut}(\Bbb Z_p)$ acts on $\Bbb Z_p$ by automorphisms. Perhaps a more popular version of this group is ${\rm Aff}(1,p)$ which is the set of matrices $$\left\{ \begin{pmatrix}k&l\\0&1\end{pmatrix}:k\in\Bbb Z_p^\times,l\in\Bbb Z_p\right\}$$ An isomorphism maps $\tau$ to $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $\sigma$ to $\begin{pmatrix}i&0\\0&1\end{pmatrix}$. This shows in particular that the group presented is independent of the choice of generator $i$.

The above gives a determination of every automorphism, and you know you can write it as $\tau^j\sigma^k$ for some $j,k<p$. In your particular case you can work out this two representatives and then understand how every other automorphism works. The group is nonabelian of order $(3-1)3=6$, so it must be $S_3$, and $\sigma,\tau$ give representatives for a transposition and a $3$-cycle, respectively. In this case you recover the usual presentation of $S_3$, namely (since $2$ generates $\Bbb Z_3^\times$ multiplcatively, and $-1=2\mod 3$)

$$\langle \sigma,\tau\mid \sigma^2=\tau^3=1,\sigma\tau\sigma^{-1}=\tau^{-1} \rangle $$


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