Showing the polynomial $f(x) = x^4 + x^3 + 4x + 1$ is irreducible in $\mathbb{Q}[x]$.
I have two question relating to this which I've bolded below.
Attempt to answer
(*) $f$ has degree $4 \ge 2$ so if $f$ has a root then $f$ is reducible.
Well the rational root test says that if $\exists$ a rational root $\frac{p}{q}$ with $p, q$ coprime then $p|a_0$ and $q|a_n$.
Hence the potential rational roots are $\pm 1$. Neither of these are roots so $f$ does not have a root. But (*) only says that if $f$ has a root then $f$ is reducible, it does not imply that if $f$ does not have a root then $f$ is irreducible. Is this correct?
Now we can't apply Eisenstein's irreducibility criterion theorem as there is no prime $p$ such that
$p\mid a_0, a_1,…, a_{n-1}$
$p \nmid a_n$
$p^2 \nmid a_0$
So where do we go from here? One other fact I am aware of is that if $f$ is primitive, which it is, $f$ irreducible in $Q[x] \iff f$ is irreducible in $Z[x]$. But I don't see how that can help me here. So how do I proceed from here?
Best Answer
Yes, that is correct. For polynomials of degree $> 3$, the absence of roots in $\mathbb{Q}$ is only a necessary, but not a sufficient condition for irreducibility.
Aside from trying a substitution $x \mapsto x - a$ to achieve a form where Eisenstein's criterion is applicable, you can consider the corresponding polynomial over $\mathbb{Z}/(p)$ for a prime $p$. If $f$ is reducible, so is its image $\overline{f} \in \left(\mathbb{Z}/(p)\right)[X]$, if it has the same degree as $f$ (since $f$ here is monic, that is the case). So if $\overline{f}$ is irreducible in any $\left(\mathbb{Z}/(p)\right)[X]$, then $f$ itself is irreducible.
Note: if $\overline{f}\in \left(\mathbb{Z}/(p)\right)[X]$ is reducible, that does not imply that $f$ is reducible.