Inverse element of $x^2 + x + 1$ in $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$

Question

I know that $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ is a field since $x^3 + x + 1$ is irreducible in $\mathbb Z_2$. But I still can't seem to find any inverse element for $x^2 + x + 1$. I want to find an element $g(x) \in \mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ such that $(x^2 + x + 1)g(x) = 1$. I've tried multiplying every single element of $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ by $x^2 + x + 1$, but I never seem to get any of them to be equal to 1 or equivalent. How is this possible if $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ is a field?

Thanks in advance.

Answer 1

Use the Euclidean algorithm to find $a,b \in \mathbb{Z}_2[x]$ so that $a(x^2+x+1)+b(x^3+x+1) = 1$. Then $a$ is the inverse.

Answer 2

Write $(x^2+x+1)(ax^2+bx+c)=1\implies ax^4+(a+b)x^3+(a+b+c)x^2+(b+c)x+c=a(-x^2-x)+(a+b)(-x-1)+(a+b+c)x^2+(b+c)x+c=(b+c)x^2+cx+c=(b+c)x^2+cx+c-a=1\implies a=1,b=0,c=0$, where I have used the fact that $x^3+x+1=0$ to substitute for $x^4$ and $x^3$.

So $x^2$ is your inverse.

Answer 3

Well, the extended Euclidean algorithm provides the solution.

I will present another way doing it:

The field $GF(8)=\Bbb Z_2[x]/\langle x^3+x+1\rangle$ has the elements

$0,1,\alpha,\alpha^2,\alpha^3=\alpha+1,\alpha^4=\alpha^2+\alpha,\alpha^5=\alpha^2+\alpha+1,\alpha^6=\alpha^2+1,\alpha^7=1,$

where $\alpha$ is a zero of $x^3+x+1$.

The inverse of $\alpha^2+\alpha+1$ you can get

(1) by recognizing that $\alpha^5=\alpha^2+\alpha+1$ and so $(\alpha^5)^{-1} =\alpha^2$, since $\alpha^7=1$,

or

(2) by solving the equation

$(\alpha^2+\alpha+1)\cdot (a\alpha^2+b\alpha +c)=1$

for $a,b,c\in\Bbb Z_2$.

Answer 4

Let $a$ be the value of $x$ modulo the ideal $(x^3+x+1)$ in $\Bbb F_2[x]$. We have to compute the inverse of $(a^2+a+1)$ in the field $\Bbb F_2(a)$. This inverse is $a^2$, quick check: $$ a^2(a^2+a+1)=a(a^3+a^2+a)=a(a+1+a^2+a)=a(a^2+1)=a^3+a=1\ . $$

Answer 5

Really, now, your field contains only eight elements, and since its multiplicative group has seven elements, it is cyclic. In other words, any element different from $1$ will generate.

Why don’t you buckle down and simply, pencil in hand, write down the successive powers of, for instance, $x$? You get $1$, $x$, $x^2$, $x^3=x+1$, et cetera. It’ll take you only a few moments to finish.