Splitting field of $x^3+x^2+1$ in $\mathbb{Z}_2$

Question

Could somebody please show me the answers and how to get to this answer? Find the splitting field of f(x)= $x^3+x^2+1$ in $\mathbb{Z}_2$ Thanks!

Answer 1

What do you know about quotient rings? The answer you are looking for is \begin{equation*} \mathbb{Z}_2[X]/(X^3+X^2+1)\cong\mathbb{Z}_2(\alpha) \end{equation*} where $\alpha$ is a root of $x^3+x^2+1$. In other words, the relation $\alpha^3+\alpha^2+1=0$ holds. Your field has the eight elements $0$, $1$, $\alpha$, $\alpha+1$, $\alpha^2$, $\alpha^2+1$, $\alpha^2+\alpha$, and $\alpha^2+\alpha+1$.

Answer 2

$f(x)$ is irreducible over $\mathbb{F}_2$ since it is a third-degree polynomial without any root in the given field.
It follows that the splitting field is simply $\mathbb{F}_8 \simeq \mathbb{F}_2[x]/(x^3+x^2+1).$