Well,
I want to show that $f(x) = x^6 – 108$ is irreducible over $\mathbb{Q}$, but I had no luck so far.
I tried to use Eisenstein criterion on $f(x+1)$ and $f(x+2)$ but it didn't work since the free coefficients doesn't suitable in both cases. I also tried to find a prime $p$ such that $f(x)\bmod p$ is irreducible, but it seems extremely inelegant.
Any other ideas?
Best Answer
The simplest thing to do is probably just to note that $\newcommand\Q{\mathbb Q}\Q(\sqrt3)\subset\Q(\sqrt[6]{108})=:L$ and $\Q(\sqrt[3]2)\subset L$. Thus, $[L:\Q]$ is divisible by $2=[\Q(\sqrt3):\Q]$ as well as by $3=[\Q(\sqrt[3]2):\Q]$, so $6\mid[L:\Q]$. At the same time, as you notices, $[L:\Q]\le6$ since it's generator satisfies a degree $6$ polynomial, so $[L:\Q]=6$ which means $x^6-108$ must be the minimal polynomial of $\sqrt[6]{108}$, and so must be irreducible.