Field $K = F_2[X]/(X^6 +X + 1)$: Show that $a^9 = a^4 + a^3$.

Question

Could someone steer me in the right direction on how to solve questions like these? Or anything I could read/watch to help me solve these kind of questions?

I have studied up on fields and know roughly what they are. I know that they're closed under addition, subtraction, multiplication, and division, but I'm not sure how to proceed.

We have a Field $K = F_2[X]/(X^6 +X + 1)$

$a$ is the class of $X$ in $K$

Show that $a^9 = a^4 + a^3$

Answer 1

Clearly we have $a^6+a+1=0$ by construction. Since in $\Bbb F_2$ we have $1+1=0$, it implies $f+f=0$ for any polynomial $f\in\Bbb F_2[X]$. Then adding $a+1$ to both sides yields $$a^6=a+1$$ Then just multiply it by $a^3$.

Answer 2

We have to show that $a^9=a^4+a^3$. $\mathbb{F}_2[x]/(x^6+x+1)$ is the field of polynomials with coefficients in $\mathbb{F}_2$ (in $\mathbb{Z}_2$, the possible coefficients of polynomials are $0$ and $1$) modulo the ideal $(x^6+x+1):=\{p(x)(x^6+x+1)|p(x) \in \mathbb{F_2}[x]\}$.

Now take $a^9$ divided by $a^6+a+1$. This gives you : $$a^9=a^4+a^3+a^3(a^6+a+1)\equiv_{\mathbb{K}}a^4+a^3$$