Irreducible Polynomial over the Rationals

Question

Question: Is $f(t) = t^4 - t + 1$ irreducible over the rationals?

My attempt: Is it valid to make a change of variables, letting $t = u + 1$? Then $f(t)$ has a root of zero, so the original polynomial is irreducible as well?

Thanks for any help!

Answer 1

Let $g$ the reduction modulo $2$ of $f$. If $x \in \mathbb{F}_4$, then $g(x) = 1$ because $x^4 = x$.

But $\mathbb{F}_2 \subset \mathbb{F}_4$, so $g$ has no roots in $\mathbb{F}_2$ and in $\mathbb{F}_4$.

And, as you know, a polynomial of degree $d$ on $\mathbb{F}_p[X]$ is irreducible if and only if it has no roots in all the $\mathbb{F}_{p^r}$ for $r \leqslant d/2$.

So $g$, of degree $4$, is irreducible over $\mathbb{F}_2$.

Conclusion : $f = X^4 - X + 1$ is irreducible over $\mathbb{Q}$.

Answer 2

Neither $1$ nor $-1$ are roots, hence $t^4-t+1$ has no linear factor. Assuming a factorization into two quadratic factors $$\begin{align} t^4-t+1&=(t^2+at+b)(t^2+ct+d)\\&=t^4+(a+c)t^3+(b+d+ac)t^2+(ad+bc)t+bd\end{align}$$ with $a,b,c,d\in\mathbb Z$ (by Gauß lemma), we find $a+c=0\Rightarrow c=-a$, $bd=1\Rightarrow b=d=\pm1$, hence $-1=ad+bc=a(d-b)=0$, contradiction.

Answer 3

$\rm g(x)=f(-x) = x^4\!+\!x\!+\!1,\,\ g(2)=19\:$ prime $\rm\Rightarrow\: g(x)$ prime, by $ $ A. Cohn's irreducibility test.