5
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $2$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Apr 24 '15 at 18:28
  • $\begingroup$ Incidentally, solving Diophantine equations algorithmically is an incredibly hard problem, so it's not exactly surprising that WA is bad at it. The general problem is just undecidable, full stop (en.wikipedia.org/wiki/Diophantine_set#Matiyasevich.27s_theorem), and even the special case of Diophantine equations in two variables is open, at least as far as I know. $\endgroup$ – Qiaochu Yuan Apr 24 '15 at 18:30
  • $\begingroup$ Ooh, and before that I forgot that the cube of a negative number is negative... Thanks for putting me straight! $\endgroup$ – sjt Apr 24 '15 at 18:31
  • $\begingroup$ 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. $\endgroup$ – Lisa Apr 25 '15 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.