Help with a finite field exercise. How to find the minimal polynomial of a given root in a given field. I need a help with this exercise.
(i) Find a primitive root $\beta$ of $\mathbb{F}_2[x]/(x^4+x^3+x^2+x+1)$. 
(ii) Find the minimal polynomial $q(x)$ in $\mathbb{F}_2[x]$ of $\beta$. 
(iii) Show that $\mathbb{F}_2[x]/(x^4 +x^3+x^2+x+1)$ is isomorphic to $\mathbb{F}_2[x]/(q(x))$.
Now, what I did for (i) was writing $\beta=x+1$ and calculating $\beta^3$ and $\beta^5$ knowing that $x^4+x^3+x^2+x+1=0$. Doing the calculation I found out that $\beta^3\neq\beta^5\neq 1$. So $\beta$ has to have order 15, hence it generates the multiplicative group of 15 elements of units of $\mathbb{F}_{16}$, and so $\beta$ is a primitive root.
The problems comes with (ii). Infact I thought that in $\mathbb{F}_2$ there is no polynomial that has $\beta=x+1$ as root, and so I can't find the asked minimal polynomial. Am I missing something? I thought that there could be a typo in the textbook and said minimal polynomial has to be found in $\frac{\mathbb{F}_2[x]}{(x^4+x^3+x^2+x+1)}$. In this case if I write $q(\alpha) = \alpha^4+\alpha^3+1$ and evaluating $q(\beta)$ (knowing that $x^4+x^3+x^2+x+1=0$) I find out that $q(\beta)=0$. Trying with polynomial of degree $<4$ doesn't give any risult. So I thought the minimal polynomial of $\beta$ is $q(\alpha) = \alpha^4+\alpha^3+1$.
Now since $q(\alpha) = \alpha^4+\alpha^3+1$ is irreducible in $\mathbb{F}_2$ (having no roots in that field), $\frac{\mathbb{F}_2}{q(\alpha)}$ is a field and has 16 elements such as the starting field. Knowing that finite field of the same cardinality are always isomorphic, (iii) is done.
Is how I resolved the exercise correct? Was I correct assuming the typo or it can be solved as the problem is stated? Can in such fields a polynomial has another polynomial as root?
Hope you can help me :)
(I don't have Galois Theory notions)
Edit: after all $q(\alpha)$ is a polynomial with coefficients in $\mathbb{F}_2$.
 A: Recall that $\Phi_5(x)=\dfrac {x^5-1}{x-1}=x^4+x^3+x^2+x+1=p(x)$.
So it seems you will want a primitive $5$-th root of unity,  $\zeta_5=e^{2\pi i/5}$.
Right,  we need to prove that the polynomial is minimal over $\Bbb F_2$.  Now it has no root.   So the question is can we factor it into two quadratics.
So write $x^4+x^3+x^2+x+1=(x^2+bx+c)(x^2+dx+f)$.  We get $(b+d)=1,(f+c+db)=1, fb+dc=1$ and $fc=1$.  Thus $f=c=1$ and $db=1\implies d=b=1$, so a contradiction,  since $b+d=1$.
You get $\Bbb F_{2^4}$ for the quotient,  because it's all polynomials in $\alpha=\zeta_5 $ of degree $\lt4$, with coefficients in $\Bbb F_2$.
Now (following you)  let's set $\beta=\alpha+1$.  Then let $q(x)=p(x-1)$.
Then  $q(\beta)=p(\alpha) =0$.
Note $\beta^5=(\alpha +1)^5=\alpha^4+\alpha =\alpha (\alpha ^3+1)\ne0$.  Remember we're in characteristic $2$.
Similarly $\beta^3=(\alpha +1)^3=\alpha ^3+\alpha ^2+\alpha +1=\alpha^4\ne0$.
Thus $\beta$ is a primitive element (order $15$).
Let's compute $p(x-1)=(x-1)^4+(x-1)^3+(x-1)^2+(x-1)+1=(x^4+1)+(x^3-x^2+x-1)+x^2+1+x=x^4+x^3+1 \pmod 2$..
This is likely the minimal polynomial of $\beta $.  There's no root.  Say $q(x)=(x^2+bx+c)(x^2+dx+f)$, then $b+d=1,f+c+bd=0,fb+dc=0, fc=1\implies f=c=1\implies 1=b+d=0$, a contradiction.
So $q(x)$ is the minimal polynomial of $\beta$.
Finally we get, by uniqueness of the field of order $16$, that $\Bbb F_2[x]/(p(x))\cong \Bbb F_2[x]/(q(x))$.
A: $x^4+x^3+x^2+x+1=\varphi_5(x)$ is a cyclotomic polynomial, hence the splitting field of such polynomial over $\mathbb{F}_2$ is $\mathbb{F}_{2^k}$ where $k$ is the least natural number such that $5\mid (2^k-1)$, i.e. the multiplicative order of $2\pmod{5}$, which is $4$ since $2$ is a generator of $\mathbb{F}_5^*$.  You have checked that the multiplicative order of $x+1$ in $\mathbb{F}_2/(\varphi_5(x))$ is $15$. In this context $x$ can be regarded as a primitive fifth root of unity $\zeta$, so the minimal polynomial of $x+1$ over $\mathbb{F}_2$ is $\varphi_5(x-1)$ reduced $\pmod{2}$, i.e. $x^4+x^3+1$. Since they differ by one $\zeta$ and $\zeta+1$ are algebraic numbers with the same degree over $\mathbb{F}_2$, and $\mathbb{F}_2/(\varphi_5(x))$ and $\mathbb{F}_2/(\varphi_5(x-1))$ are both isomorphic to $\mathbb{F}_{16}$.
