# Is $C_2$ the correct Galois Group of $f(x)= x^3+x^2+x+1$?

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?

• Something must have gone wrong in the communication between the MSE server and my brains; I'm sorry. – Lord_Farin May 31 '13 at 15:15
• 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. – DonAntonio May 31 '13 at 15:41
• @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. – Fly by Night May 31 '13 at 15:43
• @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 – Fly by Night May 31 '13 at 18:17
• @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. – Alexander Konovalov May 31 '13 at 20:27

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 $(1)(i\ {-i})$.
• 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. – Fly by Night May 31 '13 at 18:14