1
$\begingroup$

Let $\operatorname{f} \in \mathbb{Q}[x]$ where $\operatorname{f}(x) = x^3+x^2+x+1$. This is, of course, a cyclotomic polynomial. The roots are the fourth roots of unity, except $1$ itself. I get $\mathbb{Q}[x]/(\operatorname{f}) \cong \mathbb{Q}(\pm 1, \pm i) \cong \mathbb{Q}(i) = \{a+bi : a,b \in \mathbb{Q}\}.$

Let $\alpha : \mathbb{Q}(i) \to \mathbb{Q}(i)$ be a $\mathbb{Q}$-automorphism. We have: $$\alpha(a+bi) = \alpha(a)+\alpha(bi) = \alpha(a)+\alpha(b)\alpha(a)i = a+b\alpha(i).$$

Since $\alpha(i)^2 = \alpha(i)\alpha(i) = \alpha(ii) = \alpha(-1)=-1$ we have $\alpha(i) = \pm\sqrt{-1} = \pm i$. There are then two $\mathbb{Q}$-automorphisms: the identity with $\alpha(z)=z$ and the conjugate $\alpha(z)=\overline{z}$.

This tells me that the Galois Group is $S_2=\langle(12)\rangle.$ I've been using GAP software, and it says that the Galois Group is $\langle(13)\rangle$. I can see that $\langle(12)\rangle \cong \langle(13)\rangle$. However, $\langle(13)\rangle < S_3$. My suspision is that because $x^3+x^2+x+1$ is reducible over $\mathbb{Q}$: $x^3+x^2+x+1 \equiv (x+1)(x^2+1)$.

Is GAP telling me that the Galois Group of $x^3+x^2+x+1$ is $C_1\times C_2$? How should I think about the Galois Group of $x^3+x^2+x+1$? Is it $C_2$, is it a subgroup of $S_3$ which is isomorphic to $C_2$, or is it the product $C_1 \times C_2$. I realise that these are all isomorphic, but what's the best way to think of it?

$\endgroup$
  • $\begingroup$ Something must have gone wrong in the communication between the MSE server and my brains; I'm sorry. $\endgroup$ – Lord_Farin May 31 '13 at 15:15
  • $\begingroup$ In short: yes is the answer to your post's title, and $\,C_1\times C_2\cong C_2\,$ , and GAP should be executed by firing squad for writing such a nasty thing. $\endgroup$ – DonAntonio May 31 '13 at 15:41
  • $\begingroup$ @DonAntonio Gap wrote that the Galois Group is generated by $(13)$, which of course gives $C_2$. Why does is return $(13)$ and not $(12)$? My hunch is that thinking of the Galois Group as a subgroup of $S_3$ isomorphic to $C_2$ is more informative. $\endgroup$ – Fly by Night May 31 '13 at 15:43
  • 1
    $\begingroup$ @AlexanderKonovalov I was using the RadiRoot package ( gap-system.org/Packages/radiroot.html ) of GAP - Groups, Algorithms, Programming - a System for Computational Discrete Algebra gap-system.org/gap.html $\endgroup$ – Fly by Night May 31 '13 at 18:17
  • 1
    $\begingroup$ @FlybyNight: thanks, I can reproduce your example. Indeed, as RadiRoot's manual says, GaloisGroupOnRoots calculates the Galois group of f as a permutation group with respect to the ordering of the roots of f given as matrices by RootsAsMatrices. $\endgroup$ – Alexander Konovalov May 31 '13 at 20:27
0
$\begingroup$

The Galois group is the group of authomorphisms of the splitting field. It acts on the roots of any splitting polynomial (such as $f$) by permuting the roots. In your case, there are three roots, $-1, i, -i$ and the automorphisms must leave $-1$ fixed. Since the action is also free, you can view $G$ (via this action) as a subgroup of $\operatorname{Sym}(\{-1,i,-i\})$ and of cours it as only one nontrivial element $(1)(i\ {-i})$.

$\endgroup$
  • $\begingroup$ Thanks for taking the time to reply, but I already mentioned most of this in my original post. As I said, $G$ can be though of as $C_2$, as $C_1 \times C_2$ or as a subgroup of $S_3$. All of which are isomorphic. $\endgroup$ – Fly by Night May 31 '13 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.