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?


  • 10
    $\begingroup$ Why doesn't Eisenstein's criterion work? $\endgroup$ Jun 6 '14 at 8:41
  • 4
    $\begingroup$ Oh that's really annoying. So 2 divides 2, but $2^2 = 4$ doesn't divide 2... $\endgroup$
    – Wooster
    Jun 6 '14 at 8:42
  • 2
    $\begingroup$ Yes, that's correct. $\endgroup$ Jun 6 '14 at 8:42
  • 1
    $\begingroup$ We're done with Eisenstein's. Set $p=2$, $2\nmid a_n$, $2|2$, $4\nmid 2$ $\endgroup$
    – Kai
    Jun 19 '19 at 2:36

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.

  • $\begingroup$ Sorry, but I cannot see why $a$ is not an integer; could you explain more? Thanks. $\endgroup$
    – awllower
    Jun 6 '14 at 12:12
  • $\begingroup$ Let's $z=\varepsilon_{i_0}\varepsilon_{i_1}\cdots\varepsilon_{i_{k-1}}$. Only real numbers that $z$ can be are $1$ and $-1$, because $|z|=1$. On other hand $\sqrt[n]{2^{k}}$ can not be integer if $1\le k \le n-1.$ Is it clear now? $\endgroup$
    – pointer
    Jun 6 '14 at 12:28
  • $\begingroup$ Ok, it is clear now: thanks. $\endgroup$
    – awllower
    Jun 6 '14 at 12:40

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