# Why is $x^4 - 4x + 2$ irreducible over $\mathbb Q(i)$?

This is an exercise from Garling's A Course in Galois Theory Ch. 5 on irreducible polynomials.

The other part of the question was to show that $$x^5 - 4x + 2$$ was irreducible over $$\mathbb Q(i)$$. I got that one. Clearly the polynomial is irreducible over $$\mathbb Q$$ by Eisenstein and Gauss' lemma, and, since 5 and 2 are relatively prime, it follows from a previous exercise that a root of the polynomial must have degree 5 over $$\mathbb Q(i)$$. So the given polynomial must be the root's minimal polynomial over $$\mathbb Q(i)$$ (and not just over $$\mathbb Q$$). So in particular the polynomial must be irreducible over $$\mathbb Q(i)$$.

But that trick doesn't work for $$x^4 - 4x + 2$$ anymore, and I'm stumped. I convinced myself that it can't have a linear factor over $$\mathbb Z[i]$$ because that would mean a root that was a factor of $$2$$, and none of the factors of $$2$$ (namely, $$\pm 2, \pm 1, \pm 2i, \pm i, \pm (1 - i), \pm (1 + i)$$) solve the polynomial, but I don't see why it can't have a quadratic factor. Help!

• Hmm... in $\mathbb{F}_5[x]$ the polynomial $x^4 - 4x + 2$ appears to have exactly one root $x = 2$, and therefore it would be a product of a linear factor $x-2$ and an irreducible cubic factor. And also, there's a ring homomorphism $\mathbb{Z}[i] \to \mathbb{F}_5$ which sends $i \mapsto 2$. If I'm not mistaken, those facts could be put together into a proof. – Daniel Schepler Dec 11 '18 at 2:01
• I will have to think about that but thank you for the idea. – user3339517 Dec 11 '18 at 2:08
• I think that’s right, @DanielSchepler. Since this quartic has no roots in $\Bbb Q(i)$, the only possible factorization is into two quadratics. But since $(x-2)(x^3+2x^2+4x+4)$ is the only factorization into irreducibles over $\Bbb F_5$, you get a contradiction to the existence of a pair of irreducible quadratic factors over $\Bbb Q(i)$. Why don’t you write it up as an answer? – Lubin Dec 11 '18 at 3:53

By Gauss's lemma, since $$\mathbb{Z}[i]$$ is a UFD, we see that to prove $$x^4 - 4x + 2$$ is irreducible over $$\mathbb{Q}(i)$$ it will suffice to prove it is irreducible over $$\mathbb{Z}[i]$$. As you checked, this polynomial does not have any linear factors; therefore, if it were reducible, the irreducible factors would have to be quadratic.
On the other hand, over $$\mathbb{F}_5$$, the polynomial has exactly one root $$x=2$$ (which is not a double root). Therefore, here the irreducible factorization would have to be $$x^4 - 4x + 2 = (x-2) (x^3 + 2x^2 + 4x + 4)$$.
However, we have a ring homomorphism $$\mathbb{Z}[i] \to \mathbb{F}_5$$ defined by $$a + bi \mapsto a + 2b$$ (which induces an isomorphism $$\mathbb{Z}[i] / \langle 2 - i \rangle \simeq \mathbb{F}_5$$). Therefore, any factorization of $$x^4 - 4x + 2$$ into quadratic factors over $$\mathbb{Z}[i]$$ would induce a factorization into quadratic factors over $$\mathbb{F}_5$$, which is a contradiction since $$\mathbb{F}_5[x]$$ is a UFD.