I already know this is irreducible over $\mathbb{Q}$.
Right now, my general plan is to show there is no root in $\mathbb{Q\sqrt{-7}$ (thus no linear factor). And then show that two quadratic factors are impossible as well.
e,g To solve the first step, I thought about just plugging something like $a+b\sqrt{-7}$ with $a,b\in \mathbb{Q}$ into the formula to see if a root is possible but it does feel very very clunky.
Is there a better way to do this?
I am using the textbook Dummit and Foote and am working through section 14.6 on Galois Groups of Polynomials and this came up as part of solving one of the exercices (namely exercice #10, determine the Galois group of $x^4+4x-1$).
Best Answer
Edit: My original answer was wrong; the polynomial is reducible over $\Bbb{F}_9$. In fact $$x^4+4x-1=(x^2+\beta^2x+\beta)(x^2+\beta^6x+\beta^3),$$ where $\beta\in\Bbb{F}_9$ satisfies $\beta^2+\beta-1=0$. I leave my original answer here for future reference:
If the polynomial is reducible over $\Bbb{Q}(\sqrt{-7})$ then it is also reducible over its ring of integers, which is $\Bbb{Z}[\alpha]$ with $\alpha:=\tfrac12(1+\sqrt{-7})$. Then reducing mod $(3)\subset\Bbb{Z}[\alpha]$, which is prime in $\Bbb{Z}[\alpha]$, yields $$x^4+4x-1=\overline{f}\,\overline{\vphantom{f}g}\quad\text{ in }\ \Bbb{F}_9[x].$$ Of course it is not hard to factor this polynomial over $\Bbb{F}_9$, and a quick check shows that it is irreducible [Edit: It is not!]. It follows that the given polynomial is irreducible over $\Bbb{Q}(\sqrt{-7})$.