# Irreducibility of $X^n - 2$ over $\mathbf Z[i]$.

I was “playing” with Eisenstein's criterion and the polynomials $X^n-p$ ($p$ being a prime number).

Over $\mathbf Z$, the classical Eisenstein's criterion directly applies and shows that $X^n - p$ is an irreducible element of $\mathbf Z[X]$. The question was “what happens over $\mathbf Z[i]$?”

We know that in this new ring, a prime number $p$ has one of three different fates:

• if $p \equiv -1\ (\mathrm{mod}\, 4)$, $p$ still is an irreducible element of $\mathbf Z[i]$, or equivalently $(p) \subset \mathbf Z[i]$ still is prime. (Unless I'm mistaken, number theorists say that $p$ is inert). Eisenstein's criterion still applies and the polynomial is irreducible over $\mathbf Z[i]$.

• if $p \equiv 1\ (\mathrm{mod}\, 4)$, $p$ can be written as a sum of two squares: $p = n^2 + m^2$ and is therefore reducible on $\mathbf Z[i]$: $p = (n+im)(n-im)$. These two factors aren't associated (the units in $\mathbf Z[i]$ being $\pm 1$ and $\pm i$) so $(p)$ splits: $(p) = (n+im)(n-im)$ is the product of two different prime ideals. One can still use Eisenstein's criterion with either of these prime ideals, and $X^n - p$ is irreducible over $\mathbf Z[i]$.

• if $p = 2$, the oddest of all primes, one still has a factorisation $p = (1+i)(1-i)$ but this time, $1+i$ and $1-i$ are associated. In NT slang, $(2)$ is ramified: $(2) = (1+i)(1-i) = (1+i)^2$ and one cannot directly use Eisenstein's criterion.

So here's my question: how do we prove (or disprove) the irreducibility of $X^n - 2$ over $\mathbf Z[i]$?

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Let $\alpha=\sqrt[n]{2}$ then $L=\mathbb Q(\alpha)$ is a purely real extension of $\mathbb Q$. So we have that $i \notin L$ in particular $x^2+1$ is irreducible over $L$. It follows that $[L(i):L]=2$ and so $[\mathbb Q(i,\alpha):\mathbb Q(i)]=n$. Thereby $\alpha$ has degree $n$ over $\mathbb Q(i)$ so its minimum polynomial has degree $n$ which must be $x^n-2$. Since $x^n-2$ is irreducible over $\mathbb Q(i)$ it must be irreducible over its ring of integers, i.e. the Gaussian integers.
You'll notice that nothing here directly relates to $2$. We only needed that the extension $\mathbb Q(\alpha)$ was real of degree $n$. In particular this method demonstrates that if $x^n-k$ is irreducible over $\mathbb Z$ then it's irreducible over $\mathbb Z[i]$.