abstract algebra - Showing that $x^n -2$ is irreducible in $mathbb{Q}[X]$

I'm trying to show that the polynomial $X^n -2$ ($n \in \mathbb{N}$) is irreducible in $\mathbb{Q}[X]$ but am a bit stuck. Methods I know to show irreducibility:

• Gauss' lemma - which says that if I can show it is irreducible in $\mathbb{Z}[X]$ then it will be irreducible in $\mathbb{Q}[X]$.


• This would allow me to check reduction modulo a prime.

However this doesn't work in this case, because if I reduce mod 2, then the polynomial is just $X^n$ which is reducible.

I'm also aware of Eisensteins criterion where I can check that if a prime divides all the coefficients but its square doesn't divide the constant coefficient, then it is irreducible.

None of these methods are working for this polynomial so help would be much appreciated! Also are there any general methods to look out for when trying to show polynomials are irreducible?

Thanks

Answer

Let $$P(x)=x^{n}-2=(x^k+\cdots+a)(x^{n-k}+\cdots+b);~~1\le k \le n-1.$$
On other hand $$P(x)=(x-\sqrt[n]{2}\varepsilon_0)(x-\sqrt[n]{2}\varepsilon_1)\cdots(x-\sqrt[n]{2}\varepsilon_{n-1}),$$
where $\varepsilon_i=e^{\frac{2\pi i}{n}},i=0,1,\ldots,n-1.$ Therefore $$a=(-1)^{k}\sqrt[n]{2^{k}}\varepsilon_{i_0}\varepsilon_{i_1}\cdots\varepsilon_{i_{k-1}}.$$ It's easy to see that $a$ can't be integer.