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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.

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[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
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1 Answer

up vote 4 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.

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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
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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
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