Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
[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
up vote 5 down vote accepted

Having a root in the field of coefficients is equivalent to having a linear factor. If a polynomial of degree 2 or 3 factors in a non-trivial way, then at least one of the factors is linear.

share|cite|improve this answer
I want to accept your answer but I don't understand it. I will need time to stare/think at(about) your answer. – Lyapunov Jul 8 '12 at 5:52
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 $\mathbf{Q}$. – Hurkyl 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
@JyrkiLahtonen Thanks! I forgot this falls directly from a know result. – Lyapunov Jul 8 '12 at 23:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.