Determining whether ideals are principal The integral ideals of $\mathbb{Z}_{\mathbb{Q}[\sqrt{229}]}$ of norm $27$ are $\mathfrak{p}_{3}^{3}, 3\mathfrak{p}_{3}, 3\mathfrak{p}_{3}'$ and $\mathfrak{p}_{3}'^{3}$ where we write $\gamma=\frac{1+\sqrt{229}}{2}$ so that $\mathfrak{p}_{3}=(3,\gamma)$ and $\mathfrak{p}_{3}'=(3,\gamma+2)$.
How do I determine which of these are principal?
I know that $\mathfrak{p}_{3}'^{3}$ is principal since it is generated by $\beta=\frac{11+\sqrt{229}}{2}=5+\gamma$ But I have no idea about the others.
 A: Maybe you need to take a more elementary approach to this. In how many ways can 27 be expressed as a product of positive integers of degree 1? $$27 = 3 \times 9 = 3 \times 3 \times 3$$
You have already determined and confirmed (in other questions) that 3 is irreducible in this domain, but its corresponding principal ideal $\langle 3 \rangle$ is not a prime ideal, as it is properly contained in the ideals $\langle 3, 1 \pm \sqrt{229} \rangle$ (indulge me the use of $\pm$ as a notational shortcut).
In any case, the number 3 has a norm of 9. So if $N(3x) = \pm 27$, with $x \in \mathcal O_{\mathbb Q(\sqrt{229})}$, that would mean $N(x) = \pm 3$, which you already know to be impossible.
A little calculation should reveal that $(16 - \sqrt{229})(16 + \sqrt{229}) = 27$. So $\langle 16 \pm \sqrt{229} \rangle$ are principal ideals, but again, they're not prime ideals. Just after I wrote that, I noticed your $5 + \gamma$, and its conjugate $6 - \gamma$. Don't think we've found a new ideal, since $$\left(\frac{15}{2} + \frac{\sqrt{229}}{2} \right) (16 - \sqrt{229}) = \frac{11}{2} + \frac{\sqrt{229}}{2}.$$
That first multiplicand is an important number: the fundamental unit of this domain, which I like to notate $\eta$. Since $(16 - \sqrt{229}) \eta = 5 + \gamma$ and $(16 + \sqrt{229}) \overline\eta = 6 - \gamma$, this means that $\langle 16 - \sqrt{229} \rangle = \langle 5 + \gamma \rangle$ and $\langle 16 + \sqrt{229} \rangle = \langle 6 - \gamma \rangle$.
But wait a minute... factorization of ideals is unique even if factorization of numbers is not. That's kind of the point of ideals.
Therefore, if $\langle 3 \rangle = \mathfrak p \overline{\mathfrak p}$, then $\langle 27 \rangle = \mathfrak p^3 (\overline{\mathfrak p})^3$, where $\mathfrak p = \langle 3, 1 + \sqrt{229} \rangle$, and $\overline{\mathfrak p}$ is its conjugate $\langle 3, 1 - \sqrt{229} \rangle$, or vice-versa if you prefer. Or maybe you prefer $\langle 3, 1 - 2 \gamma \rangle$ and $\langle 3, -1 + 2 \gamma \rangle$, or better yet, $\langle 3, \gamma \rangle$ and $\langle 3, 1 - \gamma \rangle$, these are all equivalent, you should be able to confirm for yourself.
The point is that neither of those is principal. But of course $N(27) = 729$. So, to get an ideal of norm 27, we need to look at $\mathfrak p^3$, $(\overline{\mathfrak p})^3$, $\mathfrak p^2 \overline{\mathfrak p}$ and $\mathfrak p (\overline{\mathfrak p})^2$ (a big disadvantage of the overline notation here is that they can look like minus signs, that's the only reason I've inserted parentheses in that listing just now).
What you have to do now is multiply those ideas and determine which ones can be boiled down to principal ideals. The answers to this question should be helpful: How to multiply and reduce ideals in quadratic number ring.
And lastly, to verify your answer, figure out which two are equivalent to $\langle 16 \pm \sqrt{229} \rangle$.
