Show $p(x) = x^6 + 1.5x^5 + 3x – 4.5$ is irreducible in $\mathbb Q[x]$.
By Gauss' Lemma, a primitive polynomial in $\mathbb Z[x]$ is irreducible in $\mathbb Q[x]$ if and only if it is irreducible in $\mathbb Z[x]$. We can look at $2x^6 + 3x^5 + 6x – 9 \in \mathbb Z[x]$.
Eisenstein's Criterion fails since $3^2 \mid (-9)$. I also tried replacing $x$ with $x-1$ and $x+1$ to see if I could use Eisenstein, but they didn't work. I tried reducing it mod $p$. You cant to it mod $2$ since the leading coefficient divides 2, so I tried mod 3, but it immediately factors there. I tried mod 5 and the linear terms don't have roots, but I still need to check quadratic and cubic factors. But that just seems extremely long and if it doesn't work mod 5 I'll have to keep trying mod $p$ until I reach some prime where $p(x)$ is irreducible.
I don't know where to go from here. What is the correct way to approach this?
Best Answer
If you factor $2p$ modulo 5, then you get a product of two irreducible cubics. If you factor it modulo 7, then you get an irreducible quadratic times an irreducible quartic. This means that the polynomial is irreducible over $\mathbb{Z}$, since any factorisation over $\mathbb{Z}$ would induce a factorisation over $\mathbb{Z}/l\mathbb{Z}$ for any $l$. But the factorisations over 5 and 7 don't have a common coarser factorisation, other than the trivial one.
P.S.: No variation of the Eisenstein trick is going to work, since no prime is totally ramified in the number field $\mathbb{Q}[x]/(p)$.