$K= \mathbb{F}_2(\alpha)$ $\alpha$ root of $X^4+X+1 \in \mathbb{F}_2[X]$. Find degree and minimal polynomial Question 1: Find $[K:\mathbb{F_2}]$
Idea: I have tried looking at the irreducibility of the polynomial,  $X^4+X+1 $ and have so far been unsuccessful. Is there another way to do this apart from using Eisenstein's criterion, which I have already tried?
Then if it were irreducible, and since $\alpha$ is a root of $P(X)= X^4+X+1$, then $[\mathbb{F_2}(\alpha):\mathbb{F_2}]$ would be equal to deg$P(X)=4$
Question 2: Find the minimal polynomial for $\alpha^{-1}$ over $\mathbb{F_2}$
Idea: First find the minimal polynomial for $\alpha$ over $\mathbb{F_2}$, call this $M(X)=a_nX^n+a_{n-1}X^{n-1}+ ... + a_0$. Then the minimal polynomial for $\alpha^{-1}$ will be $N(X)=a_0 X^n+a_1 X^{n-1}+...+a_n$. Now, since $\alpha$ is a root of a degree 4 polynomial, I believe its deg$M(x)\leq 4$ although I have not managed to find it
Would really appreciate some help from you, thanks  
 A: Clearly it has no roots, so it has no irreducible linear or cubic factors. The only irreducible degree $1$ polynomial over $\Bbb F_2$ is $x^2+x+1$, so the only possible quadratic factorization is as $(x^2+x+1)^2=x^4+x^2+1$ which clearly is not your polynomial. Hence it is irreducible. This means the degree of the extension is $4$. Eisenstein is a criterion for irreducibility over the rationals though, it doesn't apply in $\Bbb F_2$
If you are worried about negative powers, then let's reformulate this as $[\Bbb F[\alpha]:\Bbb F]$ which is clearly a vector space of degree $4$ over $\Bbb F_2$. But then we know that $\Bbb F_2[\alpha]\cong \Bbb F_2[x]/(x^4+x+1)$ by the map

$$\begin{cases} \Bbb F_2[x]\to \Bbb F_2[\alpha]\\ x\mapsto \alpha\end{cases}$$

The map is clearly surjective, and by definition of the minimal polynomial and the fact that $\Bbb F_2[x]$ is a PID, we see the kernel is exactly $(x^4+x+1)$. But then the first isomorphism theorem for rings says that $\Bbb F_2[\alpha]\cong \Bbb F_2[x]/(x^4+x+1)$ as desired. Since the polynomial is generated by an irreducible in a PID, it is a maximal ideal, so the quotient $\Bbb F_2[\alpha]$ is a field. But since $\Bbb F_2(\alpha)$ is the smallest field containing $\alpha$, it must be that $\Bbb F_2(\alpha)\subseteq\Bbb F_2[\alpha]$. The inclusion in the other direction is trivial, hence the dimensions are equal.
For your second question, note that $p(x^{-1})=x^{-4}+x^{-1}+1$ is something which is zero when you put in $\alpha^{-1}$, so if you multiply it by $x^4$ you get

$$x^4p(x^{-1})=1+x^3+x^4$$

plugging in $\alpha^{-1}$ we get $\alpha^{-4}\cdot p(\alpha) = 0$, so this is the minimal polynomial for $\alpha^{-1}$, since you can get irreducibility from the same things we said for $\alpha$, namely that there are no roots, and it is not equal to $x^4+x^2+1$.
A: if $\alpha$ is a root of $P(X)= X^4+X+1$ then it exist in $\mathbb{GF_{16}}$  and indeed $\alpha$ is primitive element of $\mathbb{GF_{16}}$ .it's minimal polynomial with respect to $\mathbb{GF_2}$ is $X^4+X+1 $ itself.
for the second part we have $\alpha^{-1}$ = $\alpha^{14}$ because $\alpha^{14} = \alpha^{3}+1$ and $\alpha \times \alpha^{14} =\alpha \times (\alpha^{3}+1)= \alpha^{4}+\alpha = 1$ because we know $\alpha^{4}+\alpha+1=0$. so the minimal polynomial with respect to $\mathbb{GF_2}$ containing $\alpha^{14}$ comes from producting conjugacy class $\{ \alpha^{7},\alpha^{11},\alpha^{13},\alpha^{14} \} \Rightarrow (x+\alpha^{7})(x+\alpha^{11})(x+\alpha^{13})(x+\alpha^{14})$  which after simplification yields in:
$$x^4+x^3+1$$
