I am trying to find the ring of integers of $\mathbb{Q}(\sqrt[3]{6})$. Unfortunately, the discriminant of the basis $\{1, \theta, \theta^2\}$, where $\theta = \sqrt[3]{6}$, is equal to $2^2 (-3)^5$ (so it's not square-free). How can I proceed in this case to find the ring of integers? I could take the prime $2$, and analyze the expressions of the form $(a + b\theta + c\theta^2)/2$ for binary values of $a,b,c$, but maybe there is a more direct approach.
Ring of integers of $\mathbb{Q}(\sqrt[3]{6})$
abstract-algebraalgebraic-number-theoryextension-field
Best Answer
$Disc(x^3-6)= -27(-6)^2=3^a2^b$
Since $v_3(6^{1/3})=v_3(6)/3=1/3$ then $\Bbb{Q}_3(6^{1/3})/\Bbb{Q}_3$ is totally ramified of degree $3$ with uniformizer $6^{1/3}$
Similarly $\Bbb{Q}_2(6^{1/3})/\Bbb{Q}_2$ is totally ramified with uniformizer $6^{1/3}$
The uniformizers of the ramified completions are in $\Bbb{Z}[6^{1/3}]$ thus it is a Dedekind domain ie. $$O_{\Bbb{Q}(6^{1/3})} = \Bbb{Z}[6^{1/3}]$$ For $p\ne 2,3$ then $(p)$ is a product of distinct prime ideals of $\Bbb{Z}[6^{1/3}]$, for $p=2,3$ then $(p)=(p,6^{1/3})^3$.
The more elementary solution : check that $(p)=(p,6^{1/3})^3$ for $p=2,3$, since the unramified prime ideals are easily shown to be inversible, this implies every prime ideal is inversible, thus it is a Dedekind domain, thus it is integrally closed ($=O_K$). The point is that (in Dedekind domains) a prime ideal $P$ becomes principal $=(\pi)$ in $(O_K-P)^{-1} O_K$, the uniformizer is $\pi$, from which we obtain a discrete valuation and a $p$-adic completion.