# $f(x)$ is irreducible if and only if $f(x)$ does not have a root in $\mathbb{Z}/\mathbb{2Z}.$

Let $f(x)$ be a polynomial in $(\mathbb{Z}/\mathbb{2Z})[x]$ of degree $2$ or $3$. Prove that $f(x)$ is irreducible if and only if $f(x)$ does not have a root in $\mathbb{Z}/\mathbb{2Z}.$

I know that $f(x)$ is irreducible if and only if $F[x]/(f(x))$ is a field.

Any suggestions/hints will be appreciated.

• [abstract-algebra] is a tag best used with other tags; your latest questions have all been about finite fields. Why not add the [finite-fields] tag, then? – Arturo Magidin Jul 8 '12 at 5:43
• @ArturoMagidin Will do from now onwards. I didn't know that tag existed! I learned Abstract algebra from three different professors (and books) since undergrad days, so I am confused about notations and conventions! – Lyapunov Jul 8 '12 at 5:48

• If $f(x)=(ax+b)g(x)$, with $a\neq0$ and $g(x)$ another polynomial, then $f(-b/a)=0$. If $a$ and $b$ are in some field, then so is $-b/a$. – Jyrki Lahtonen Jul 8 '12 at 5:58
• Note that the field $\mathbf{Z} / 2 \mathbf{Z}$ has nothing to do with it. The proof over that field is exactly the same as the proof over $\mathbf{Q}$. – user14972 Jul 8 '12 at 6:05
• +1 to Hurkyl's comment. It is easier to check for presence of roots in $\mathbf{Z}/2\mathbf{Z}$ though :-) – Jyrki Lahtonen Jul 8 '12 at 6:07