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

$a^9+a^4+a^3$ is divided by $a^6+a+1$. Also, $-1=1$ in $\mathbb F_2$. — Vercassivelaunos, Oct 24 '20 at 15:55
@Vercassivelaunos Thanks but I would love to get a more thorough explanation on this. I'm still not sure what's going on exactly. — SJ19, Oct 24 '20 at 16:02
You need to start utilizing mathjax when you post questions, SJ19. — amWhy, Oct 24 '20 at 16:11
The answerers got it right. You could also do the following. You know that $a^6+a+1=0$. Therefore $a^6=-a^6=a+1$. Therefore $a^7=a\cdot a^6=a(a+1)=a^2+a$. Thereforet $a^8=a(a^7)=a(a^2+a)=a^3+a^2$. Therefore...? There will be more terms, when the exponent gets higher, but as long as you remember that $(1+1)=0$ and $1=-1$ in the field $K$ you will do fine. — Jyrki Lahtonen, Oct 24 '20 at 16:53
If you think you need more exercises on this type of problems, try and reproduce the tables in this answer that I prepared for referrals like this. — Jyrki Lahtonen, Oct 24 '20 at 16:55

score 2 · Accepted Answer · answered Oct 24 '20 at 16:03

2

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 $a^3$.

answered Oct 24 '20 at 16:03

Berci

90,745

Thanks, but I think I'm missing some fundemental knowledge on this. If you don't mind could you explain more about how classes works? The meaning of "a" especially. – SJ19 Oct 24 '20 at 16:54
We quotient out by the equivalence relation for which $f\sim g$ iff $g-f$ is in the given ideal $(X^6+X+1)$, that is, if $g-f$ is a multiple of $X^6+X+1$. Thus, in particular $X^6+X+1\sim 0$. $a$ is the equivalence class of $X$. Taking the quotient can also be thought as simply regarding equals the equivalents, basically replacing $\sim$ in the ambient set to $=$ in the quotient set. Because we work with an ideal, all the ring operations are well defined in the quotient set. – Berci Oct 24 '20 at 17:37

score 2 · Answer 2 · answered Oct 24 '20 at 16:27

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 : $$a^9=a^4+a^3+a^3(a^6+a+1)\equiv_{\mathbb{K}}a^4+a^3$$

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

2 Answers2