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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!

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    $\begingroup$ 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. $\endgroup$ – Daniel Schepler Dec 11 '18 at 2:01
  • $\begingroup$ I will have to think about that but thank you for the idea. $\endgroup$ – user3339517 Dec 11 '18 at 2:08
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    $\begingroup$ 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? $\endgroup$ – Lubin Dec 11 '18 at 3:53
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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.

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