Prove roots of $x^5 - 1$ are constructible I am trying to show some other result, and by reducing it to this problem I should be able to finish the proof, I am trying to show that $z = e ^ {\frac{2}5  \pi  i} $ is constructible (In the sense that there exists a chain of field extensions of degree 2 from $\mathbb{Q}$ that include $z$)
I already found the minimal polynomial of $z$ : $f = x^4 + x^3 + x^2 + x + 1$ as I showed in general that this is irreducible. Now the only issue I am having is that, for $F = \mathbb{Q}[x] / (f)$, we have that the degree of the field extension is $[F : Q] = 4$ (the degree of $f$). However I am not so sure on how should I proceed from this, I think I should express this field extension as two field extensions of degree two but I haven't figured out how to do this step.
I have searched online and saw that it's enough to show that the splitting field of f has degree $2^N$ in order to show the constructibility of its roots, but I haven't found a proof of this.
EDIT: After studying Galois theory the result follows from the fact that the splitting field of a separable polynomial is always a Galois extension (and $|Gal(F/\mathbb Q)| = [F : \mathbb Q]$), thus by Sylow theorems you can get subgroups for every order $2^k$, and by the Fundamental Theorem of Galois Theory this proves the existence of all the intermediate fields you need.
Another approach I thought about would be to just use the construction of the pentagon, but I wanted to understand better the problem in terms of abstract algebra.
 A: Hint
Since the equation $x^4+x^3+x^2+x+1=0$ is symmetric, the standard substitution $t=x+\frac{1}{x}$ solves it. Note that at this point you already know that $x$ is constructible: the substitution leads to an equation of degree $\frac{4}{2}=2$ isn $t$ with rational coefficients, and hence $t$ is constructible. Moreover, $x$ is the root of the quadratic $x^2-tx+1=0 \in \mathbb Q(t)[X]$ and thus constructible.
But, just to see here are the details
$$t=x+\frac{1}{x}\\
t^2-2=x^2+\frac{1}{x^2}$$
Then 
$$x^4+x^3+x^2+x+1=0 \Rightarrow x^2+x+1+\frac{1}{x} +\frac{1}{x^2}=0 \Rightarrow x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0 \Rightarrow \\ t^2-2+t+1=0 \Rightarrow t^2+t-1=0$$
Just solve for $t$, and then solve
$$t=x+\frac{1}{x}\Rightarrow x^2-tx+1=0$$
A: Since $[F : \mathbb Q] = 4$, it suffices to find an intermediate extension $ \mathbb Q \subset K \subset F$.
Then $[K : \mathbb Q] = 2$ and $[F : K] = 2$ implies that they are constructible.
As mentioned in the comments, $K=\mathbb Q(z+\bar z)$ is a natural candidate.
Indeed, let $w = z+\bar z$. Since $\bar z=z^4$, we have $wz=z^2+1$ and thus this is the minimal equation for $z$ over $K=\mathbb Q(w)$.
A: Given a fourth degree equation with integer coefficients, all roots are constructible iff the resolvent cubic to this equation has a rational root.  The roots of the quartic are then constructible functions of this rational resolvent root.
Let's drop this condition on $x^5-1=0$.  First factor out $x=1$ so our fourth-degree equation is $x^4+x^3+x^2+x+1=0$.
$x^4+x^3+x^2+x+1=[x^2+[(1/2)+2\sqrt{s}]x+t_1][x^2+[(1/2)-2\sqrt{s}]x+t_2]$
Match terms with like powers:
$x^4,x^3:$  trivial.
$x^2:  t_1+t_2+[(1/4)-4s]=1, t_1+t_2=(3/4)+4s$
$x^1:  t_1[(1/2)-2\sqrt{s}]+t_2[(1/2)+2\sqrt{s}]=(t_1+t_2)/2-2(t_1-t_2)\sqrt{s}=1, t_1-t_2=[s-(5/16)]/\sqrt{s}$
$x^0:  4t_1t_2=(t_1+t_2)^2-(t_1-t_2)^2=4, s(3/4+4s)^2-[s-(5/16)]^2=4s, \color{blue}{4096s^3+1280s-720s-25=0}$
The blue equation represents the resolvent cubic, which may be rendered as $(16s)^3+5(16s)^2-45(16s)-25=0$ to simplify the search for rational roots.
We identify $s=5/16$ as a root and with that, constructibility is proved.  The rational root for $s$ leads to constructible values of $t_1$ and $t_2$, thus constructible quadratic factors for our fourth-degree equation.  Just like Euclid and Ptolemy said there would be when they derived constructions for the regular pentagon.
