Prove that $ x^4 - 2 $ is irreducible over $ \mathbb{Z}[i] $

Question

How do I prove that $ p(x) = x^4 - 2 $ is irreducible over $ \mathbb{Z}[i] $?

This seems very elementary yet I'm not sure how to do it.

Someone suggested using Eisenstein and $ p = 1+i $, but this doesn't seem right because $ (1+i)^2 = 2i $ is an associate of $ -2 $.

I have seen somewhere that one can use a generalized version of the Rational Root Theorem and simply check that $ 1+i $ and $ 1-i $ are not roots of $ p(x) $, is this correct?

Thank you for your help.

Answer 1

Checking roots is insufficient, as the equation had degree $4$ so it could potentially factored as 2 degree 2 polynomials.

Reduce the equation mod (3+2i). Then $\mathbb{Z}[i]\rightarrow\mathbb{F}_{13}$, $i\rightarrow 5$, and any integer get reduced mod 13=(3+2i)(3-2i). Then the equation is now $X^{4}-2$. If the equation was reducible before, it is still reducible now. But if it is reducible now, either it has a root, or it breaks as product of quadratic factors so either way all of its root lie in $\mathbb{F}_{169}$. Then $\mathbb{F}_{169}^{\times}$ is a cyclic group of order $168$, and $2$ has order $12$, so any roots must have order $48$ which is impossible as $48$ does not divide $168$.

EDIT: same calculation will also work for $\mathbb{F}_{5}$, as the other answer suggested.

Answer 2

In this case I would prefer the following way which is also alluded in the comments:

$x^4-2=(x-\sqrt[4]{2}) (x+\sqrt[4]{2}) (x-i\sqrt[4]{2}) (x+i\sqrt[4]{2}) $ in $\mathbb{C}$. Obviously $\sqrt[4]{2}, -\sqrt[4]{2}, i\sqrt[4]{2}, -i\sqrt[4]{2}$ are not in $\mathbb{Z}[i]$. Since $\sqrt[4]{2}\cdot \sqrt[4]{2}=\sqrt{2}\notin \mathbb{Z}[i]$, there is no product of polynomials of degree two in $\mathbb{Z}[i]$.

Answer 3

You can also note that $x^4-2$ is irreducible over $\mathbb F_5$ and $\mathbb F_5 = \mathbb Z[i]/(2+i)$.