Galois Group of $p(x)=x^{5}+x^{4}+1$ For fun, I’m trying to find the Galois group of the polynomial $p(x)=x^{5}+x^{4}+1$.
I solved for the roots and got the splitting field $\mathbb{Q}\left(\sqrt[3]{\frac{9+\sqrt{69}}{8}}+\sqrt[3]{\frac{9-\sqrt{69}}{8}}, i\sqrt{3}\right)$. I now need to find the automorphism group of this, but I don’t know where to start.
I know an automorphism maps some nth root to its conjugate, but I don’t know what to do with such a complicated field extension.
 A: I guess it behooves me to promote the comments to an answer.

The quintic factors as
$$
f(x)=x^5+x^4+1=(x^3-x+1)(x^2+x+1).
$$
That particular quadratic factor is a friend we meet often. See for example this old answer for a description of a general class of trinomials having it as a factor. The cubic factor is gotten by polynomial division. Both factors are seen to be irreducible in $\Bbb{Q}[x]$ because we can quickly exclude the possibility of rational zeros (use the rational root theorem). As the factors have degrees $\le3$, this implies their irreducibility.
Let $\omega=e^{2\pi i/3}=(-1+i\sqrt3)/2$ be a primitive complex root of unity of order three. The zeros of the quadratic factor are then $z_4=\omega$ and $z_5=\overline{\omega}=\omega^2$.
We then use Cardano's method for finding the complex zeros of the cubic factor. Following the procedure outlined e.g. here, if $u^3$ and $v^3$ are the roots of
$$
y^2+y+\frac1{27}=0\qquad(*)
$$
such that $uv=1/3$, then $u+v$ is a zero of $x^3-x+1$. By the quadratic formula the roots of $(*)$ are $(-9\pm\sqrt{69})/18$. If $\alpha=\root3\of{(-9+\sqrt{69})/18}$ and $\beta=\root3\of{(-9-\sqrt{69})/18}$ are the real cube roots (both negative), then we have $\alpha\beta=1/3$, and Cardano's formula gives
$$
\begin{aligned}
z_1&=\alpha+\beta,\\
z_2&=\omega\alpha+\omega^2\beta,\\
z_3&=\omega^2\alpha+\omega\beta
\end{aligned}
$$
as the roots of the cubic factor $x^3-x+1$.
The splitting field of $f$ is thus $L=\Bbb{Q}(z_1,z_2,z_3,z_4,z_5)$. Because $z_5=z_4^2$ we can dispose of $z_5$ as a generator. Because $z_1+z_2+z_3=0$ (Vieta) we can also dispose of $z_3$. Hence $L=\Bbb{Q}(z_1,z_2,z_4)$. Because the factor $x^2+x+1$ brought the primitive third roots of unity into $L$, we see that
$\alpha=(z_1-\omega z_2)/(1-\omega^2)\in L$. Therefore $L=\Bbb{Q}(\alpha,\omega)$.
The primitive element theorem further states that we can actually get $L$ by adjoining a single element to $\Bbb{Q}$. However, that is not very helpful for the purposes of identifying the Galois group $G=Gal(L/\Bbb{Q})$. I prefer to describe it as a group of permutations of the set of roots $X=\{z_1,z_2,z_3,z_4,z_5\}$.
If $\sigma\in G$ is some automorphism of $L$, then it must permute the elements of $X$ and once its action of $X$ is known, it will be completely determined. Obviously $\sigma$ must permute $z_4$ and $z_5$ amongst themselves (2 alternatives) and $z_1,z_2,z_3$ amongst themselves (6 alternatives). That gives a total of $2\cdot6=12$ possibilities. It turns out that all those $12$ permutations come from actual automorphisms in $G$, but that is not yet totally clear. To that end we calculate the discriminant, $D$, of the cubic. By plugging in the numbers we get $D=-23$. We observe that $D$ is not the square of a rational number. This has the following consequences (known from basic Galois theory):

*

*The splitting field $K$ of the cubic factor $x^3-x+1$ is a degree six extension of $\Bbb{Q}$. Furthermore, the Galois group $Gal(K/\Bbb{Q})$ is isomorphic to $S_3$. In other words, it permutes the zeros $z_1,z_2,z_3$ in all the possible ways.

*The only quadratic subfield of $K$ is $\Bbb{Q}(\sqrt{D})=\Bbb{Q}(\sqrt{-23})$.

*Because this subfield does not coincide with the splitting field $F=\Bbb{Q}(\omega)$ of the quadratic factor, we see that $K\cap F=\Bbb{Q}$.

*As both $K/\Bbb{Q}$ and $F/\Bbb{Q}$ are Galois, the previous bullet implies that they are linearly disjoint. It follows that their compositum $L=KF$ satisfies
$$[L:\Bbb{Q}]=[K:\Bbb{Q}]\cdot[F:\Bbb{Q}],$$ and
$$G\simeq Gal(K/\Bbb{Q})\times Gal(F/\Bbb{Q})=S_3\times C_2$$
with the projections from $G$ to the two factor groups coming from restricting the automorphisms of $L$ to $K$ and $F$ respectively.

There are otherwise of seeing that $[L:\Bbb{Q}]=12$, and that already implies that $|G|=12$, and hence all the described $12$ permutations of $X$ must represent automorphisms.

If we look at $G$ by how it acts on the pair of generators $\alpha$ and $\omega$ we get a slightly different view. The zeros of
$$g(x)=x^6+x^3+\frac1{27}$$
are $\omega^i\alpha$ and $\omega^i\beta$, $i=0,1,2$. Let $M=\Bbb{Q}(\alpha)$. As
$$12=[L:\Bbb{Q}]=[L:M]\cdot[M:\Bbb{Q}]=[M(\omega):M]\cdot[M:\Bbb{Q}],$$
it follows that $[M:\Bbb{Q}]$ must be equal to six (clearly $[M(\omega):M]=2$). Therefore $g(x)$ must be the minimal polynomial of $\alpha$. Hence $\alpha$ has the listed six roots of $g(x)$ as its conjugates. We can conclude that $L$ is also the splitting field of $g(x)$.
