I am given that for the ring of integers of $K = \mathbb{Q}(\sqrt[3]{5})$ is $\mathcal{O}_K = \mathbb{Z}[\sqrt[3]{5}]$. I am supposed to factorise the ideals $(2), (3), (5)$ and $(7)$, show that all prime ideal factors are principal and then use Minkowski's bound to deduce that $\mathcal{O}_K$ is a PID, however I am having problems with $(2)$ and can't work out what I am doing wrong!
Given the above assumption we have that $[\mathcal{O}_K : \mathbb{Z}[\sqrt[3]{5}]] = 1$ so that we can apply Dedekind's Theorem for all primes $p$. Doing this for $p = 2$ gives:
$$x^3 - 5 \equiv x^3+1 \equiv (x+1)(x^2+x+1) \tag{mod 2} $$
So that we have $(2) = \mathfrak{p}_2\mathfrak{p}_4 = (2, \sqrt[3]{5} + 1)(2, \sqrt[3]{25} + \sqrt[3]{5} + 1)$
Then I think I should show that $\mathfrak{p}_2$ and $\mathfrak{p}_4$ are principal by finding elements in $\mathcal{O}_K$ with norms 2 and 4 respectively. I calculated the norm to be:
$$\text{Norm}(a+b\sqrt[3]{5}+c\sqrt[3]{25}) = a^3 + 5b^3 + 25c^3 - 15abc $$
But when I put this equal to 2 or 4 in WolframAlpha I don't get any results in $\mathbb{Z}[\sqrt[3]{5}]$. Thus I want to argue that $\mathfrak{p}_2$ and $\mathfrak{p}_4$ are not principal ideals and so $\mathcal{O}_K$ is not a PID.
What went wrong?
