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Since $x^5 - 4x - 2$ is irreducible (using Eisenstein's criterion with $p=2$), and it has precisely three real roots, it follows that the Galois group of the splitting field is $S_5$. Therefore, there exists a quadratic extension of $\mathbb{Q}$ contained in this field, corresponding to the normal subgroup $A_5$. In fact, since the sign map $S_5 \to \{ \pm 1 \}$ is an abelianization map, $A_5$ should be the unique (normal) subgroup of $S_5$ of index 2, so the quadratic extension should be unique. So, I found myself wondering how one would calculate exactly what field this is.

I suppose a brute force method would be: if the roots are $\alpha, \beta, \gamma, \delta, \epsilon$, then this splitting field has basis $\{ \alpha^{0..4} \beta^{0..3} \gamma^{0..2} \delta^{0..1} \epsilon^0 \}$ as a $\mathbb{Q}$-vector space using the relations: \begin{align*} \alpha^5 - 4 \alpha - 2 & = 0 \\ \beta^4 + \alpha \beta^3 + \alpha^2 \beta^2 + \alpha^3 \beta + (\alpha^4 - 4) & = 0 \\ \gamma^3 + (\alpha + \beta) \gamma^2 + (\alpha^2 + \alpha \beta + \beta^2) \gamma + (\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3) & = 0 \\ \delta^2 + (\alpha + \beta + \gamma) \delta + (\alpha^2 + \alpha \beta + \beta^2 + \alpha \gamma + \beta \gamma + \gamma^2) & = 0 \\ \epsilon + (\alpha + \beta + \gamma + \delta) & = 0. \end{align*} Then $\frac{1}{60} \operatorname{Tr}^L_K = \frac{1}{60} \sum_{\sigma \in A_5} \sigma \cdot$ would be a projection of $L$ onto $K$ if $L$ is the splitting field and $K$ is the quadratic field. Therefore, some basis element would have irrational trace; and then we could calculate the (quadratic) irreducible polynomial of the image of this basis element.

However, I was wondering if there might be some other method to calculate the answer which doesn't require nearly as much work. The problem is, if I try taking a simple basis element like $\alpha^1$, then the trace of that is zero; the norm would be $\alpha^{12} \beta^{12} \gamma^{12} \delta^{12} \epsilon^{12} = 2^{12}$; etc. In fact, I think the only basis element with irrational trace would be $\alpha^4 \beta^3 \gamma^2 \delta$ since otherwise, there's some transposition which fixes it and so $\operatorname{Tr}^L_K(x) = \frac{1}{2} \operatorname{Tr}^L_{\mathbb{Q}}(x) \in \mathbb{Q}$.

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  • $\begingroup$ Hmm, whatever it is, it must be a complex extension: the subgroup corresponding to $L \cap \mathbb{R}$ is generated by a transposition of the two complex roots, so it's not contained in $A_5$, so $L \cap \mathbb{R}$ does not contain the quadratic field. $\endgroup$ – Daniel Schepler Dec 11 '17 at 20:17
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The quadratic extension is $\Bbb Q(\sqrt D)$ where $D$ is the discriminant of the polynomial $x^5-4x-2$. As the polynomial is a trinomial, computing its discriminant won't be too hard.

See here for a formula to find this discriminant.

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  • $\begingroup$ OK, now I feel silly. Anyway, my computation (based on numerical approximations of the roots, but the determinant formula should also work) shows that $D = -2^4 \cdot 13259$ so the quadratic field is $\mathbb{Q}(\sqrt{-13259})$. $\endgroup$ – Daniel Schepler Dec 11 '17 at 20:49
  • $\begingroup$ And Wolfram Alpha agrees on the value of the discriminant. $\endgroup$ – Daniel Schepler Dec 11 '17 at 20:52

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