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I'm trying to draw the subgroup diagram for the Galois group of $x^3 - 4x + 2$, and find the subfield fixed by each subgroup.

Here's what I know: the group is $S_3$. I solved the equation through Vieta's method, and found that a root was:

$$ t = \frac{4}{\sqrt{3}} \cos\left(\frac{1}{3} \arccos\left(\frac{3\sqrt{3}}{-8}\right)\right) $$

Through checking the discriminant I found that the splitting field was $\mathbb{Q}(\sqrt{37}, t)$. But how do I determine of this messy field? I think that

$$\mathbb{Q}, \mathbb{Q}(t, \sqrt{37}) \mathbb{Q}(\sqrt{37}), \mathbb{Q}(t), \mathbb{Q}(t\sqrt{37}), \mathbb{Q}(t^2\sqrt{37})$$

will be the subfields, but how do I know which subgroup fixes which one? I'm having a hard time thinking of a concrete example of an automorphism of the roots, because $t$ is so messy.

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You are right that $\Bbb Q(t)$ and $\Bbb Q(\sqrt{37})$ are intermediate subfields, but it turns out that $t\sqrt{37}$ and $t^2\sqrt{37}$ are primitive elements, generating the entire field. You can check this by noting that they aren't fixed by any nonidentity element of $S_3$. (Note that the 2-cycle fixing $t$ sends $\sqrt{37}$ to $-\sqrt{37}$, being an odd permutation).

It's easier to connect fixed fields with subgroups in this case by viewing the action of $S_3$ on the three roots of your polynomial $t, t_2, t_3$, where $t_2, t_3$ are the other two roots of $x^4-4x+2$. The three subgroups given by 2-cycles act by switching two roots and leaving the third root fixed. It follows (by consideration of degree, for instance) that the corresponding intermediate fields are given by $\Bbb Q(t), \Bbb Q (t_2), \Bbb Q(t_3)$. The only remaining intermediate field corresponds to the subgroup $A_3$ and is given by $\Bbb Q(\sqrt{37})$.

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