# Is $\mathbb{Z}(\sqrt[3]{5})$ a PID? Factorisation of the ideal $(2)$

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?

What went wrong is that you trusted WolframAlpha! Set $c = 0$ and then $b = 0$ to simplify the norm and you'll quickly find that $\sqrt[3]{5} - 1$ has norm $4$, while $3 - \sqrt[3]{25}$ has norm $2$.
• I don't know how WolframAlpha attempts to solve Diophantine equations, but checking a few examples it seems to know how to solve linear equations and Pell equations and that's about it. It told me that $y^2 = x^3 + 9$ has no integer solutions, but of course $(3, 6)$ is a solution. – Qiaochu Yuan Apr 24 '15 at 18:28
• Wolfram Alpha is excellent in my opinion. But you have to remember that it's a tool, not an oracle. Tools have their limitations and kludges. Kind of like trying to figure out the complex cubic roots of $8$ using a cash register. It can be done, but you're going to need a lot of workarounds. – Lisa Apr 25 '15 at 21:21