<s>[INST] Find Primitive Root for Polynomial Field Can someone help me get started on the problem below:
Recall that $\mathbf{F}_{p^k}$ can be realized as $\mathbf{F}_p[x]/P(x)
  \cdot \mathbf{F}_p[x]$ where $P(x)$ is a polynomial of degree $k$ with
  coefficients in $\mathbf{F}_p$ which is irreducible.
Problem Find primitive roots for the fields $\mathbf{F}_8
  \cong \mathbf{F}_2[x]/(x^3+x+1) \cdot \mathbf{F}[x]$ (with elements $a+bx+cx^2$
  for $a,b,c \in \mathbf{F}_2$) and for $\mathbf{F}_9 \cong \mathbf{F}_3[x]/(x^2+1)$ (with
  elements $a+bx$ for $a,b \in \mathbf{F}_3$).  To do it, you just need to
  show that some polynomial has multiplicative order equal to $7$ in
  the first case and $8$ in the second case.
 [/INST] In the quotient $\mathbf{F}_8 \cong \mathbf{F}_2/(x^3+x+1)$ you have the relation
$$x^3+x+1=0 \iff x^3=-x-1=x+1,$$
since over $\mathbf{F}_2, -1=1$. You can use this relation to perform multiplications. In this case $\mathbf{F}_8^{\times} \cong \mathbf{Z}/7\mathbf{Z}$. By Lagrange the order of an element divides the order of the group, so it divides $7$, which is prime in $\mathbf{Z}$, therefore it is either $1$ or $7$, but the only element which has order $1$ is the unit $1$, so every other element has order $7$. 
We can prove this directly, so that you get comfortable with how to use the relation above. Let us take $x \in \mathbf{F}_8$. (Again, you can take whatever element you like, so, if you want, just try it) We show that $\langle x \rangle =\mathbf{F}_8^{\times}$.
$x^1 = x;$ 
$x^2 =x^2;$
$x^3 = x+1;$
$x^4 = x\cdot x^3 = x(x+1)=x^2+x;$
$x^5 = x^2\cdot x^3 = x^2(x+1)=x^3+x^2=x+1+x^2;$
$x^6 = x^3\cdot x^3 = (x+1)^2 = x^2+2x+1=x^2+1,$ since over $\mathbf{F}_2, 2=0$;
$x^7 = x^6 \cdot x = (x^2+1)x = x^3+x = x+1+x=1.$
I let you do the other case. Just notice that the relation it holds this time is $x^2+1=0$...
If you have problems with it just ask :-).
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