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I am asked to deduce, in the following order, 3 facts about $G = Gal(f(t) = t^6+3)$ over the rationals:

  1. $G$ is order 6
  2. The elements have orders $1,2,3$
  3. $G \cong S_3$

I have figured these out but in the wrong order so I wanted to write down a different train of thought that I am less sure of to see if there is anything wrong with it.

  1. By Eisenstein $f(t)$ is irreducible, and is thus the minimal polynomial of a root $\beta$ of $f$ over $\mathbb Q$, so if $L$ is the splitting field of $f$ over $\mathbb Q$ then as the degree of $f$ is $6$, then the degree of the extension $\mathbb Q\leq L$ is also $6$.

  2. Letting $\xi$ be a primitive $6$-th root of unity, then we see that over $\mathbb Q(\xi) = \mathbb Q(i\sqrt 3), \; f$ factorises as $f(t) = (t^3 + i\sqrt 3)(t^3 - i\sqrt 3)$ . This means that all permutations $\sigma \in Gal(L/\mathbb Q(\xi))$ permute the roots of each cubic factor amongst themselves. i.e. $\sigma$ is a $3$-cycle, and there are three of them. Also we note that complex conjugation preserves $\mathbb Q$ in $L$, so $G$ contains an order $2$ permutation. The identity is order $1$. I am not quite sure how to conclude that there is no element of order $6$, but with that we have that the elements are of orders $1,2,3$

  3. Finally, we notice that composing complex conjugation with one of the $3$-cycles gives us the inverse of the original $3$-cycle. This gives a dihedral relation and hence $G$ is the dihedral group of order $6$, i.e. $G \cong S_3$

I would like to clarify why it is exactly that we can conclude that there is no element of order $6$ in $G$. I can feel that it is a very simple thing that I have overlooked, but I can't figure out what it is and would appreciate a little help or a hint. Additionally I would really appreciate it if anyone could let me know if there's anything flawed with my logic here at all. Thank you.


marked as duplicate by Jyrki Lahtonen Dec 29 '17 at 6:23

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You have basically the right idea but there are some details you haven't stated exactly right. Here is how I would clean up your argument.

First, in your step 1, you have only found that $[K:\mathbb{Q}]=6$, not that $[L:\mathbb{Q}]=6$. This tells you $|G|=[L:\mathbb{Q}]\geq 6$. Equality does follow from some of your later work though.

Let's now look at step 2. To justify your assertions about $Gal(L/\mathbb Q(\xi))$ more precisely, let's let $\alpha$ be a chosen cube root of $i\sqrt{3}$. Then from your factorization of $f$, we see that the roots of $f$ are $\alpha$, $\alpha\xi^2$, $\alpha\xi^4$, $-\alpha$, $-\alpha\xi^2$, and $-\alpha\xi^4$. This proves that $L=\mathbb{Q}(\alpha,\xi)=\mathbb{Q}(\alpha)$ since $\xi\in\mathbb{Q}(\alpha^3)$, and so proves that in the previous paragraph, $K=L$ and $G$ really does have only $6$ elements. We also see that $[L:\mathbb{Q}(\xi)]=[L:\mathbb{Q}]/[\mathbb{Q}(\xi):\mathbb{Q}]=3$, so $Gal(L/\mathbb Q(\xi))$ can only be a cyclic group of order $3$. Since $L$ is the splitting field of $t^3-i\sqrt{3}$ over $\mathbb{Q}(\xi)$, this cyclic group of order $3$ must permute the cube roots of $i\sqrt{3}$ cyclically, as you claimed. (But there are only two such elements of order $3$; the third element is the identity!)

So at this point, we have found two elements of order $3$ and one element of order $1$ in $G$ (forming a cyclic subgroup of order $3$). You then correctly observe that complex conjugation is in $G$, giving an element of order $2$.

At this point I don't see any direct way to conclude $G$ has no elements of order $6$. However, if $G$ had an element of order $6$, it would be cyclic, so it suffices to show $G$ is not abelian. So, let's look at your step 3 now. You claim that composing complex conjugation with one of the 3-cycles gives its inverse, but that is false! Rather, conjugating one of the 3-cycles by complex conjugation gives its inverse.

To show this, let's write everything down explicitly in terms of $\alpha$ and $\xi$. There is a generator $\sigma\in Gal(L/\mathbb{Q}(\xi))$ such that $\sigma(\alpha)=\alpha\xi^2$. If $\tau$ denotes complex conjugation, we have $\tau(\xi)=\xi^{-1}$. To find $\tau(\alpha)$, note first that $\tau(\alpha)^3=\tau(i\sqrt{3})=-i\sqrt{3}$, so $\tau(\alpha)$ is one of $-\alpha$, $-\alpha\xi^2$, and $-\alpha\xi^4$. Note also that $\alpha$ is not purely imaginary, so $\tau(\alpha)$ cannot be $-\alpha$. We thus may assume $\tau(\alpha)=-\alpha\xi^2$ (if not, then change which primitive 6th root we're calling $\xi$ to make it true).

Now we can compute the compositions of $\sigma$ and $\tau$ explicitly. We have $$\sigma(\tau(\alpha))=\sigma(-\alpha\xi^2)=-\sigma(\alpha)\xi^2=-\alpha\xi^4$$ and $$\tau(\sigma(\alpha))=\tau(\alpha\xi^2)=-\alpha\xi^2\xi^{-2}=-\alpha.$$

In particular, $\sigma\tau\neq\tau\sigma$, so $G$ is nonabelian and has no elements of order $6$. At this point you can conclude $G\cong S_3$ since $S_3$ is the unique nonabelian group of order $6$ up to isomorphism. Alternatively, you can use similar computations to those above to find that in fact $\tau\sigma\tau=\sigma^{-1}$ and get the dihedral relation you referred to and an explicit isomorphism to $S_3$.

  • $\begingroup$ Thank you this! I'll be reading through this later today more thoroughly, but I noticed a mistake I made in part 1. Id written $K \leq L$ but I meant to put $\mathbb Q \leq L$. In your response, what is the field $K$ that you have where you say that $[K:\mathbb Q] = 6$? $\endgroup$ – user366818 Dec 29 '17 at 11:51
  • $\begingroup$ Sorry I'm also struggling to see how you conclude from the fact that the roots are $\alpha, \xi^2 \alpha, \xi^4 \alpha, -\alpha, -\xi^2 \alpha, -\xi^4 \alpha$ that $\xi \in \mathbb Q(\alpha^3)$ $\endgroup$ – user366818 Dec 29 '17 at 12:33
  • $\begingroup$ Er, by $K$ I meant the field $\mathbb{Q}(\beta)$ obtained by adjoining one root. A priori, this is not the same as $L$, since adjoining just one root may not give you the entire splitting field. $\endgroup$ – Eric Wofsey Dec 29 '17 at 15:57
  • $\begingroup$ Those are the roots since the roots are the cube roots of $i\sqrt{3}$ and $-i\sqrt{3}$, and $\xi^2$ is a primitive cube root of unity. So $\alpha,\xi^2\alpha,\xi^4\alpha$ are the cube roots of $i\sqrt{3}$ and their negatives are the cube roots of $-i\sqrt{3}$. $\endgroup$ – Eric Wofsey Dec 29 '17 at 15:58
  • $\begingroup$ Ah I understood that they were the roots, I was confused how you concluded that $\xi \in \mathbb (\alpha^3)$ Im sorry I must be missing something really obvious. $\endgroup$ – user366818 Dec 29 '17 at 22:58

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