I have the polynomial $x^3 -2 \in \mathbb{Q}[x]$, of which the splitting field is $E = \mathbb{Q}(2^{1/3}, \omega)$ where $\omega = e^{2\pi i/3}$. The Galois group of $E$ over $\mathbb{Q}$ is isomorphic to $S_3$, and I wish to find the intermediate fields $\mathbb{Q} \subset K \subset E$.

So $S_3$ has $4$ proper non-trivial subgroups, namely $A_3$, $\{id, (1,2)\}$, $\{id, (1,3)\}$ and $\{id, (2,3)\}$. Here is where I'm stuck.

I decided to look at $\mathbb{Q}(2^{1/3})$ first. Since $\mathbb{Q}(2^{1/3}) = \{a+b2^{1/3}+c4^{1/3}\}$, and both $2^{1/3}$ and $4^{1/3}$ have separable minimal polynomials over $\mathbb{Q}$ I initially thought that $\mathbb{Q}(2^{1/3})$ is a finite separable extension which allows me to look at the fields of invariants and use a strong result in my book. Also, $[\mathbb{Q}(2^{1/3}):\mathbb{Q}] \leq 3$. However, I couldn't quite find the automorphisms of that Galois group, so maybe it's not supposed to be $3$? I'm just not exactly sure how to solve this kind of problem, and some help would be nice.

  • $\begingroup$ $\mathbb Q(2^{⅓})$ is separable (in fact every finite extension of a characteristic $0$ field is separable) but is not normal! Otherwise, it would be the splitting field of $X^3-2$. $\endgroup$ – Mathmo123 May 27 '16 at 11:57
  • $\begingroup$ @Mathmo123 Yes, I completely misread the result in my book. So then $[\mathbb{Q}(2^{1/3}):\mathbb{Q}] \leq 3$. I could only find $2$ automorphisms in the Galois group, so are these all there is? $\endgroup$ – Auclair May 27 '16 at 11:59
  • $\begingroup$ Well the degree of the extension is $3$ so there should be $3$ automorphisms in the Galois group. Since $\mathbb Q(2^{⅓})$ is not a Galois extension, you would not expect these automorphisms to preserve $\mathbb Q(2^{⅓})$. Each of the automorphisms acts by sending $2^{⅓}$ to one of the three roots of $X^3-2$. $\endgroup$ – Mathmo123 May 27 '16 at 12:06

Any finite extension of a field of characteristic $0$ is separable, so in particular $\Bbb{Q}(2^{⅓})\supset\Bbb{Q}$ is. It is however not normal, as otherwise it would be the splitting field of its minimal polynomial $X^3-2$, which you already know it isn't.

To identify all intermediate fields $\Bbb{Q}\subset K\subset E$ using Galois theory, in stead of looking at $\Bbb{Q}(2^{⅓})$ or any other intermediate field, you should look at intermediate groups $\{\operatorname{id}_E\}\subset H\subset\operatorname{Gal}(E/\Bbb{Q})$. You have identified all four nontrivial intermediate groups, and they correspond one-to-one to the intermediate fields; each group corresonds to the subfield of $E$ that it fixes.

So how does $S_3$ act on $E$? How does $\{\operatorname{id}_E,(2\ 3)\}$ act on $E$, for example, and what subfield is fixed?

  • 1
    $\begingroup$ That's what I did after my first approach. I started looking at $A_3$ and tried to identify the elements of $E$ which were invariant under the automorphisms of $A_3$. I haven't finished it yet, but thank you for the answer. $\endgroup$ – Auclair May 27 '16 at 12:25

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