# splitting over Z mod 2

• April 13th 2009, 05:19 PM
frankdent1
splitting over Z mod 2
let $\alpha$ be a zero of $f(x)=x^3+x^2+1$ over $\mathbb{Z}_{2}$, Show that $f(x)$ splits over $\mathbb{Z}_{2}(\alpha)$

I don't understand this question because $f(x)$ is irreducible in $\mathbb{Z}_{2}$ ( $f(0)=1$ and $f(1)=1$) so how can $\alpha$ be its zero.
• April 13th 2009, 08:05 PM
whipflip15
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).
• April 13th 2009, 09:31 PM
NonCommAlg
Quote:

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

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

as whipflip15 suggested: show that if $\alpha$ is a zero of $x^3 + x^2 + 1 \in \mathbb{Z}_2[x],$ then: $x^3+x^2+1=(x-\alpha)(x-\alpha^2)(x-\alpha^4).$