Given $\langle X^6+X^3+1\rangle \in \mathbb{F}_2[X]$ is an irreducible polynomial, let's define $R:= \mathbb{F}_2[X]/ \langle X^6+X^3+1 \rangle$
How much elements does R contain?
$R$ contains $2^6=64$ elements
How many elements does the set of units contain?
Here, I know that every non zero element is either a zero divisor or a unit, but since $X^6+X^3+1$ is irreducible, $R$ is a field and therefore there are no zerodivisor, but still, how do I know how many elements?
- What is the multiplicative order of the element $X + \langle X^6+X^3+1\rangle$?
I was thinking about reasoning as follow: (but not sure exactly what to do)
$X^0=1$
$X^1=X$
$X^2=X^2$
$X^3=X^3$
$X^4=X^4$
$X^5=X^5$
$X^6=[-X^3-1]$
...
Any clue for this point, please ?
EDIT:
$X^6=[-X^3-1]$
$X^9=X^3[-X^3-1]=[-X^6-X^3]=-1$
$X^{15}=-X^6=[X^3+1]$
$X^{18}=X^3[X^3+1]=[X^6+X^3]=1$