# Showing a polynomial $f\in\mathbb Q[x]$ is irreducible if it has rational coefficients?

I'm trying to figure out how I can do this for some arbitrary function. Say I find a monic associate of $f$ that we'll call $f_1(x)$. If I then apply Eisenstein's Criterion or Descartes' Rational Root Test to $f_1(x)$ and find it's irreducible, what does this say about $f(x)$? Is there a theorem for this?

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That it is irreducible too. It is like $7$ and $-7$ are both irreducible in $\mathbb{Z}$. –  1015 Apr 17 '13 at 2:43

## 1 Answer

Hint $\rm\ f\:$ reducible $\Rightarrow$ $\rm\:uf\:$ reducible, since $\rm\:f = gh\:$ $\Rightarrow$ $\rm\:uf = (ug)h.\:$ Thus, contrapositively, we deduce that $\rm\:uf\:$ irreducible $\rm\Rightarrow\: f\:$ irreducible (here $\rm\:u\:$ is any unit).

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Ok I see. Thanks –  Caleb Jares Apr 17 '13 at 2:47