The Eisenstein integers are $\mathbb Z[\omega]$ where $\omega$ is the primitive third root of unity. If $K$ is some algebraic number field, can $\mathcal O_K$ be isomorphic to $\mathbb Z[\omega]$?
Attempt: The rank of $\mathcal O_K$ as a $\mathbb Z$-module is the degree of the extension $K$ of $\mathbb Q$. I believe that $\mathbb Z[\omega]$ has rank 2, as it has $\mathbb Z$-basis $1$, $\omega$, since $\omega^2=-\omega -1$ which is in the $\mathbb Z$-linear span of $1$, $\omega$. (is this a correct argument?). Hence $K$ must be a quadratic extension - $K=\mathbb Q(\sqrt d)$. Depending on the congruences of $d$ mod $4$, $\mathcal O_K\cong \mathbb Z[\sqrt d]$ or $\mathbb Z\left[\frac{1+\sqrt d}{2}\right].$ $\omega =\frac{-1+\sqrt{-3}}{2}=-1+\frac{1+\sqrt{-3}}{2}$, so am I correct in saying that $\mathbb Z[\omega]=\mathbb Z\left[\frac{1+\sqrt d}{2}\right]$ with $d=-3?$ And now luckily $-3\equiv 1 \mod 4,$ so indeed if $K$ is $\mathbb Q(\sqrt{-3}),$ $\mathcal O_K\cong \mathbb Z[\omega].$ (I apologise if this seems simple, but I'm a bit unfamiliar with $\mathbb Z$-modules, so I don't know if manipulations like these are justified.)
Follow-up question: If the above is correct, is this some special case of a result like "If $\xi_n$ is a primative $n$th root of unity then $\mathbb Z[\xi_n]$ is isomorphic to $\mathcal O_K$ for some $K$ and if $n$ is prime, $[K:\mathbb Q]=n-1?$" This seems to work with $n=4$ too when, we get the Gaussian integers and (more trivially) if $n=1,2$ we get $\mathbb Z$, the ring of integers of $\mathbb Q$.
If $\zeta_n$ is any primitive $n$th root of unity and $K=\mathbb{Q}(\zeta_n)$ (this $K$ is known as the $n$th cyclotomic field) then we have $\mathcal{O}_K=\mathbb{Z}[\zeta_n]$ and $[K:\mathbb{Q}]=\varphi(n)$, where $\varphi$ is the Euler phi function. A good reference for this would be chapter 2 of Marcus' Number Fields.
In particular, you're correct that the Eisenstein integers $\mathbb{Z}[\zeta_3]$ are the ring of integers of the number field $\mathbb{Q}(\zeta_3)=\mathbb{Q}(\sqrt{-3})$.
You've also correctly observed that $\varphi(n)=n-1$ when $n$ is prime.