Element order is not a divisor

Question

I have field $\mathbb F_2 [\alpha]$, where $\alpha^{6} = \alpha^4 +\alpha^3 + \alpha + 1$. As I know, this is field, because $p(x)=x^6+x^4+x^3+x+1$ irreducible polynomial (if ain't wrong). So, i need to find ord($\alpha$). As I know, possible ord($\alpha$) = {1, 3, 7, 9, 21, 61}, because this numbers are dividers of |$\mathbb F_2[\alpha]^{*}$| = 63.

But ord($\alpha$)=20, which is not dividers 63.

Could it be?

Computing: $\alpha^{6} = \alpha^4 +\alpha^3 + \alpha + 1$ => $\alpha^{9} = \alpha^{6} \cdot \alpha^{3} = \alpha^{3} \cdot (\alpha^4 +\alpha^3 + \alpha + 1) = \alpha^5 +\alpha^4 + \alpha^2 + 1 $

$\alpha^7 = \alpha^5 + \alpha^4 + \alpha^2 + \alpha$

$\alpha^8 = \alpha^7 \alpha = \alpha^6 + \alpha^5 +\alpha^3 + \alpha = \alpha^5 + \alpha^4 + 1$

$\alpha^{10} = \alpha^9 \alpha = \alpha^5 + \alpha^4 +1$

$\alpha^{20} = (\alpha^{10})^2 = \alpha^{10} + \alpha^8 + 1 = \alpha^5 + \alpha^4 +1 + \alpha^5 + \alpha^4 +1 + 1 = 1$

Answer 1

We have $$ p(x)=x^6-x^4-x^3-x-1=(x^3 + x^2 + 2x + 1)(x^2 + 1)(x + 2) $$ with $p(\alpha)=0$, so it is not irreducible over $\Bbb F_3$.

Edit: Over $\Bbb F_2$ it is indeed irreducible. Then see for example this related post: Factoring $x^6 + x^4 + x^3 + x + 1$ over $\mathbb{F}_{16}$

How did you compute the order of $\alpha$? My calculations are different (let me write $a$ for $\alpha$): \begin{align*} a^7 & = a^5+a^4+a^2+a,\\ a^8 & = a^5+a^4+a^2+a+1,\\ a^9 & = a^5+a^4+a^2+1,\\ a^{10} & = a^5+a^4+1,\\ a^{11} & = a^5 + a^4 + a^3 + 1,\\ a^{20} & = a^2+a+1,\\ a^{21} & = a^3+a^2+a,\\ a^{63} & = 1. \end{align*} So the order is $63$, and the problem has dissolved.