Let $L=\mathbb{F_2}(\theta)$ where $\theta$ is a root of $X^4+X+1$
I am trying to solve the following consecutive questions in Galois Theory:
Prove that $L$ has degree $4$ over $\mathbb{F_2}$ and contains a primitive $5$-th root of unity
Need to show $[\mathbb{F_2}(\theta) : \mathbb{F_2}]=4$. This is the case if the minimal polynomial of $\theta$ over $X^4+X+1$, which I believe is true.
I am not sure how to show that $L$ contains a primitive $5$-th root of unity.
Apply Krummer's theory to find an irreducible polynomial of degree $5$ in $L[X]$
This polynomial is of the form $X+aX^4+bX^3+cX^2+dX+e$ and has no linear, quadratic or cyclic factors.
Is it true that $a, b, c, d, e \in \{0, 1, \theta\}$
How could we use Kummer's theory here?
The polynomial $X^4+X+1$ is irreducible in $\mathbf F_2[X]$ since it has no zeros in $\mathbf F_2$, and it is not the square of the unique irreducible polynomial of degree $2$ in $\mathbf F_2[X]$. It follows that $X^4+X+1$ is the minimal polynomial of $\theta$ over $\mathbf F_2$. Since $L=\mathbf F_2[\theta]$, the degree of the extension $L/\mathbf F_2$ is $4$, i.e., $L=\mathbf F_{16}$.
In particular, the multiplicative group $L^\times$ is cyclic of order $16-1=15$. One has
$$ \theta^3\neq1\quad\text{and}\quad\theta^5=\theta\cdot\theta^4=\theta(\theta+1)=\theta^2+\theta\neq1 $$ since $1,\theta,\theta^2,\theta^3$ is an $\mathbf F_2$-basis of $L$. In particular, $\theta$ is a generator of $L^\times$.
The $5$-th roots of unity in $L$ are then $1,\theta^3,\theta^6,\theta^9,\theta^{12}$, and are generated by $\theta^3$.
Anyway, the polynomial $X^5-\theta$ does not have any root in $L$, since the $5$-th powers in $L$ are $0,1,\theta^5,\theta^{10}$ which are all different from $\theta$.
By Kummer theory, $X^5-\theta$ is irreducible over $L[X]$. In fact, that is easy to see. Let $M=L[\eta]$, with $\eta$ a root of $X^5-\theta$. Then, $\eta\not\in L$ by what has been said above. Multiplying $\eta$ by any $5$-th root of unity gives again a root of $X^5-\theta$ in $M\setminus L$. This means that $M$ is the splitting field of $X^5-\theta$ over $L$. Since $L$ is perfect, $M/L$ is Galois. Since $M\supsetneq L$, there is at least $1$ nontrivial element $\sigma$ in the Galois group of $M/L$. One has $$ \sigma(\eta)=(\theta^3)^i\eta, $$ for some $i\in\{1,2,3,4\}$. Since $i$ is relatively prime with $5$, the element $\sigma$ acts transitively on the roots of $X^5-\theta$ in $M$. It follows that $X^5-\theta$ is irreducible in $L[X]$.