What about $\mathbb{Z}[\sqrt[n]{2}]$? Most textbooks talk about $\mathbb{Z}[\sqrt{d}]$ but I haven't succeed in my search of direct references about $\mathbb{Z}[\sqrt[n]{d}]$.
I would like to ask about $\mathbb{Z}[\sqrt[n]{2}]$ for $n>2$. I know it inherits the norm from $\mathbb{Q}(\sqrt[n]{2})$, but what is it known about $\mathbb{Z}[\sqrt[n]{2}]$? Is it a euclidean domain, or a unique factorization domain, or at least a GCD domain? Can their prime or irreducible elements be clasiffied as nicely as in $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{-1}]$? I will be very grateful with answers or references for this questions, even for just $\mathbb{Z}[\sqrt[3]{2}]$.
Thank you.
 A: The discriminant of $\alpha=\sqrt[3]{2}$ is $\pm 27\cdot 4$, so only $2$ and $3$ may ramify, since $\pm 27\cdot 4=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2 d_K$, where $K=\Bbb Q\!\left(\sqrt[3]2\right)$.
Now $\mu_\alpha=X^3-2$ is $2$-Eisenstein, so $2\nmid  [\mathcal{O}_K:\mathbb{Z}[\alpha]]$ and $\mu_{\alpha}(X-1)$ is $3$-Eisenstein, so  $3\nmid  [\mathcal{O}_K:\mathbb{Z}[\alpha +1]]=  [\mathcal{O}_K:\mathbb{Z}[\alpha]]$. Therefore,  $\mathcal{O}_K=\mathbb{Z}[\alpha]$ and $d_K=\pm 27\cdot 4$. It follows that the factorisation of $(p)$ into prime ideals is reflected by the decomposition of $X^3-2$ modulo $p$.
The  Minkoswki bound is $\approx 2.94$, hence the class group is generated by the class of prime ideals lying above $2$. Now, since $(2)=(\alpha,2)^3=(\alpha)^3$, the unique prime ideal above $2$ is $(\alpha)$, which is principal, so $\mathbb{Z}[\alpha]$ is a PID. A conjecture (supported by lot of positive results) suggests that a ring of integers is a PID if and only it is Euclidean (but not necessarily norm Euclidean, so the Euclidean function may be very hard to find).
Since $\mathbb{Z}[\alpha]$ is a PID, irreducible elements are exactly the generators of prime ideals. Depending of the factorisation of $X^3-2$ mod $p$, you will have.
For $p\neq 2,3$:
If $2$ is not a cube modulo $p$, $p$ is irreducible
If $2$ is a cube modulo $p$ but $X^2+X+1$ is irreducible modulo $p$, we have $(p)=P_1P_2$, where $P_1=(\alpha-m,p), P_2=(\alpha^2+m\alpha+m^2,p)$n where $m^3\equiv 2 \ [p]$.
The generators $\pi_1,\pi_2$ of $P_1,P_2$ will give you non associate irreducible elements.
If $2$ is a cube modulo $p$ and $X^2+X+1$ is reducible modulo $p$, whe have $(p)=P_1P_2P_3$, where $P_k=(\alpha-j^km,p)$ with $j^2+j+1\equiv 0 \ [p]$ and $m^3\equiv 2 \ [p]$.
The generators $\pi'_1,\pi_2',\pi'_ 3$ of $P_1,P_2,P_3$ will give you non associate irreducible elements.
If $p=2$, we get that $\alpha$ is irreducible
If $p=3$, $(3)=P^3$, where $P=(\alpha-1,3)=(\alpha-1)$ , hence $\alpha-1$ is irreducible.
Hence, up to association, the irreducible elements are $\alpha,\alpha-1$ and the various $\pi_1,\pi_2,\pi'_1,\pi'_2,\pi'_3$. I doubt that you can find the last ones explicitely for general $p$...
A: According to LMFDB - The L-functions and Modular Forms Database:

*

*$\Bbb Z\!\left[\sqrt[3]2\right]$is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[4]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[5]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[6]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[7]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[8]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[9]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[10]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[11]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[12]2\right]$ is a Principal Ideal Domain.

*$\Bbb Z\!\left[\sqrt[20]2\right]$ is a Principal Ideal Domain assuming GRH.

*$\Bbb Z\!\left[\sqrt[40]2\right]$ is a Principal Ideal Domain assuming GRH.

There must be something fishy going on...
Methodology: the link says that an integral basis is $1, a, a^2, \cdots$ which means that $\Bbb Z\!\left[\sqrt[n]2\right]$ is the ring of integers of $\Bbb Q\!\left(\sqrt[n]2\right)$, and it also says that the class number is $1$, which means that the ring is a Principal Ideal Domain.
