# Thread: splitting over Z mod 2

1. ## splitting over Z mod 2

let $\displaystyle \alpha$ be a zero of $\displaystyle f(x)=x^3+x^2+1$ over $\displaystyle \mathbb{Z}_{2}$, Show that $\displaystyle f(x)$ splits over $\displaystyle \mathbb{Z}_{2}(\alpha)$

I don't understand this question because$\displaystyle f(x)$ is irreducible in $\displaystyle \mathbb{Z}_{2}$ ($\displaystyle f(0)=1$ and $\displaystyle f(1)=1$) so how can $\displaystyle \alpha$ be its zero.

2. The zero does not lie in Z_2. It lies in an extension of Z_2 containing alpha.

To help solve problem note that alpha^2 is also a root in Z_2(alpha).

3. Originally Posted by frankdent1
let $\displaystyle \alpha$ be a zero of $\displaystyle f(x)=x^3+x^2+1$ over $\displaystyle \mathbb{Z}_{2}$, Show that $\displaystyle f(x)$ splits over $\displaystyle \mathbb{Z}_{2}(\alpha)$

I don't understand this question because$\displaystyle f(x)$ is irreducible in $\displaystyle \mathbb{Z}_{2}$ ($\displaystyle f(0)=1$ and $\displaystyle f(1)=1$) so how can $\displaystyle \alpha$ be its zero.
as whipflip15 suggested: show that if $\displaystyle \alpha$ is a zero of $\displaystyle x^3 + x^2 + 1 \in \mathbb{Z}_2[x],$ then: $\displaystyle x^3+x^2+1=(x-\alpha)(x-\alpha^2)(x-\alpha^4).$