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I'm stuck between two possibilities for a Galois group of an extension, and I could use a hint on how to decide.

First we consider the Galois group of the splitting field of $h(X) = X^3 + 3X + \sqrt 3\in \mathbb Q(\sqrt3)[X]$ over $\mathbb Q(\sqrt 3)$ (that $h$ is prime and seperable is easy to see). Observe that if $\alpha$ is a root of $h$ then the minimal polynomial of $\alpha$ is $f(X) = X^6 - 6X^4 + 9X^2 - 3$, meaning that $[\mathbb Q(\alpha):\mathbb Q] = 6$. The discriminant of $h$ is $\Delta(h) = 27 = (3\sqrt3)^2$ i.e. it is a square in $\mathbb Q(\sqrt3)$. By multiplicativity of degree, this forces $\operatorname{Gal}(h/\mathbb Q(\sqrt 3))\simeq C_3$.

The last part in fact means that $\mathbb Q(\alpha)/\mathbb Q(\sqrt 3)$ is Galois, which in this case (not in general) implies that $\mathbb Q(\alpha)/\mathbb Q$ is Galois (in both cases the condition that needs checking is normality).

The goal is to calculate $\operatorname{Gal}(\mathbb Q(\alpha)/\mathbb Q)$.

A further observation is that the Galois group of the Galoisian closure of $\mathbb Q(\alpha^2)/\mathbb Q$ is also $C_3$, which, since we already know $[\mathbb Q(\alpha) : \mathbb Q] = 6$, means that $\operatorname{Gal}(\mathbb Q(\alpha)/\mathbb Q(\alpha^2))\simeq C_2$.

So there is a subgroup of order $2$, and a subgroup of order $3$. If there are more subgroups of order $2$, the group we're trying to calculate is $S_3$. If not, it is $C_6$. This is a practice exam question, which means there should be something I can do that isn't just loading up Maple and calculating roots (which I could do since $X^6 - 6X^4 + 9X^2 - 3 = (X^3 + 3X + \sqrt 3)(X^3 + 3X - \sqrt 3)$). However it isn't coming to me.

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A hint mostly.

Note that if $a$ is a root of $$x^3+x+\sqrt{3}=0$$ then $-a$ is a root of $$x^3+x-\sqrt{3}=0$$

Thus permutations of the roots of these two polynomial are the same. Now use this to check if the Galois group is commutative.

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  • $\begingroup$ Thanks! This means the roots are $\{a,b,c,-a,-b,-c\}$. Then if $\sigma$ has order three wlog $\sigma = (a,b,c)$ and the group is commutative iff the transposition is $\tau = (b,c)$. But this is impossible since in that case $\tau (-b) = -c$. Yes? $\endgroup$
    – GPerez
    Dec 9, 2016 at 2:14
  • $\begingroup$ Not quite, yes you have $\sigma=(a,b,c)(-a,-b,-c)$ and it fixes $\sqrt{3}$. You now need $\tau$ such that $\sqrt{3}\mapsto -\sqrt{3}$. How does $\tau$ act on the roots ? You can multiply by a power of $\sigma$ is necessary and assume that $\tau(a)=-a$. It follows then by looking at the discriminant that $\tau(b)=-b$, etc. So $\tau$ is $(a,-a)(b-b)(c,-c)$. Now its easy to see that $\sigma$ and $\tau$ commute. $\endgroup$ Dec 9, 2016 at 2:24
  • $\begingroup$ Ah, I made a mistake in writing the commutativity in terms of the roots, but I understand now. $\endgroup$
    – GPerez
    Dec 9, 2016 at 3:03
  • $\begingroup$ Sorry, I think I spoke to soon. I can write down the discriminant of $f$ as $(8abc(a^2-b^2)(a^2-c^2)(b^2-c^2))^2$. How should this help me decide $\tau(b)$, for instance? (I understand up to $\tau(a)=-a$) $\endgroup$
    – GPerez
    Dec 9, 2016 at 3:42
  • $\begingroup$ I think you have $b-c=\frac{3\sqrt{3}}{(a-b)(a-c)}$ if I understand your calculation in the question. Now the denominator is symmetric in $b$ and $c$ so the value of $\tau$ is determined there. Thus you have $\tau(b-c)=c-b$ and the only way to have that is $\tau(b)=-b$. $\endgroup$ Dec 9, 2016 at 3:47

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