# Deducing Galois group of $t^6 + 3$ over the rationals [duplicate]

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 LahtonenDec 29 '17 at 6:23

• I'm sure this question has come up here before. I'll have a look.... – Gerry Myerson Dec 28 '17 at 18:25
• – Gerry Myerson Dec 28 '17 at 18:26

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 $S_3$.

• 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 $[K:\mathbb Q] = 6$? – user366818 Dec 29 '17 at 11:51
• 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 $\xi \in \mathbb Q(\alpha^3)$ – user366818 Dec 29 '17 at 12:33
• 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. – Eric Wofsey Dec 29 '17 at 15:57
• 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 $-i\sqrt{3}$. – Eric Wofsey Dec 29 '17 at 15:58
• 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. – user366818 Dec 29 '17 at 22:58