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I'm trying to find the Galois group of the polynomial $p(X)= X^5+X+1$ over $\mathbb Q$.

First, one notes that, if $\omega$ is a primitive cubic root of unity, then it is a root of $p(X)$. So, $X^2+X+1$ is a factor of $p(X)$. We have the following factorisation: \begin{equation} p(X) = (X^2+X+1)(X^3-X^2+1). \end{equation} Now, the derivative of $p(X)$ is $p'(X) = X^4+1$, which is always positive. Hence, $p(X)$ has only one real root, which is a root $\alpha$ of $X^3-X^2+1$. Let me denote by $\eta$ and $\overline{\eta}$ the other two complex roots of $X^3-X^2+1$. Then, a splitting field of $p(X)$ over $\mathbb Q$ is clearly given by \begin{equation} K = \mathbb Q(\alpha, \omega, \eta). \end{equation} Next, I have to compute the degree $[K : \mathbb Q]$ of the extension. It is simple to prove that $[\mathbb Q(\alpha, \eta) : \mathbb Q] = 6$: in fact, $\mathbb Q(\alpha, \eta)$ is a splitting field of $X^3-X^2+1$ over $\mathbb Q$, and this polynomial has only one real root. So, I try to compute as follows: \begin{equation} [K : \mathbb Q] = [K : \mathbb Q(\alpha,\eta)][\mathbb Q(\alpha, \eta) : \mathbb Q]. \end{equation} Here comes my problem: clearly, the degree $[K : \mathbb Q(\alpha,\eta)]$ is $1$ or $2$. The point is to understand whether $\omega \in \mathbb Q(\alpha, \eta)$ or not, and I'm unable to give an answer. I think that actually $[K : \mathbb Q(\alpha,\eta)] = 2$, and that eventually one can prove that the Galois group $\mathrm{Gal}(K | \mathbb Q)$ is isomorphic to $S_3 \times C_2$.

How can I prove this?

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  • $\begingroup$ See math.stackexchange.com/a/38898/589 and math.stackexchange.com/questions/45893/…. $\endgroup$ – lhf Jun 6 '14 at 16:30
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    $\begingroup$ Nicely presented and worked out question. +1 $\endgroup$ – DonAntonio Jun 6 '14 at 16:32
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    $\begingroup$ Can anyone explain why for the splitting field we only need to adjoin $\eta$? I don't see how we can generate $\overline\eta$ from that. $\endgroup$ – Jack M Jun 6 '14 at 16:38
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    $\begingroup$ $\overline{\eta}$ is the complex conjugate of $\eta$. Adding $\eta$ to $\mathbb Q(\alpha)$ is the same as adding the root of the discriminant of the quadratic equation whose roots are $\eta$ and $\overline{\eta}$, which is the same as adding $\overline{\eta}$. $\endgroup$ – Francesco Genovese Jun 6 '14 at 16:47
  • $\begingroup$ Related: math.stackexchange.com/questions/1375747 $\endgroup$ – Watson Dec 26 '16 at 13:07
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The only quadratic subextension of $\Bbb Q \subset \Bbb Q(\alpha,\eta)$ is given by the subfield fixed by the index $2$ subgroup $A_3$ of $S_3$. It is generated by $(\alpha-\eta)(\eta-\bar{\eta})(\bar{\eta}-\alpha)$, which is the square root of the discriminant $\Delta$ of $X^3-X^2+1$

A computation gives $\Delta = 23$. Since $\Bbb Q(\sqrt{23})$ and $\Bbb Q(\sqrt{-3})$ are different extensions, $\sqrt{-3} \notin \Bbb Q(\sqrt {23})$ and so $\sqrt{-3} \notin \Bbb Q(\alpha,\eta)$

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