# Ring of integers of $\mathbb{Q}(\sqrt{6})$

I am trying to find the ring of integers of $$\mathbb{Q}(\sqrt{6})$$. Unfortunately, the discriminant of the basis $$\{1, \theta, \theta^2\}$$, where $$\theta = \sqrt{6}$$, 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.

$$Disc(x^3-6)= -27(-6)^2=3^a2^b$$

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.

• I do not understand this solution; in my class we have not talk about uniformizer. Do you know a less technical approach different that the one I mentioned? – Philomeno Dec 6 '19 at 3:03
• In general, if the minimal polynomial of $\theta$ is Eisenstein with respect to the prime $p$, then the index of $\Bbb Z[\theta]$ in the ring of integers is coprime to $p$. If you know about extensions of the field of $p$-adic numbers this is straightforward, but maybe is less easy to prove from first principles. @Philomeno – Angina Seng Dec 6 '19 at 3:09
• @LordSharktheUnknown I feel better already =). – Philomeno Dec 6 '19 at 3:17

If you are interested, a complete determination of the ring of integers $$R$$ of a pure cubic field $$\mathbf Q(\sqrt  m)$$, where $$m$$ is a cube free integer written as $$m=hk^2$$, withe $$h,k$$ coprime and square free, can be found in chap.3 of D.A. Marcus' book "Number Fields".

In the general case of a number field $$\mathbf Q(\alpha)$$, where $$\alpha$$ is an algebraic integer of degree $$n$$, thm.13 therein states that $$R$$ admits an integral basis of the form {$$1, f_1(\alpha)/d_1,..., f_{n-1}(\alpha))/d_{n-1}$$}, where the $$d_i \in \mathbf Z$$ , with $$d_1|d_2|...|d_{n-1}$$, are uniquely determined, and the $$f_i \in \mathbf Z[X]$$ are monic of degree $$i$$. In the particular case of $$\alpha=\sqrt  m$$, everything becomes explicit. The (long) exercise 41 op. cit. shows that one can take $$f_1(\alpha)/d_1=\alpha$$, and $$f_2(\alpha)/d_2=\alpha^2/k$$ if $$m\neq \pm 1$$ (mod $$9$$), $$\alpha^2\pm k^2\alpha+k^2/3k$$ if $$m\equiv \pm 1$$ (mod $$9$$) .

• For $h,k$ coprime squarefree the same argument as in my answer shows that $\Bbb{Z}[3^{-1},(hk^2)^{1/3},(h^2k)^{1/3}]$ is a Dedekind domain, thus it remains to find the uniformizer of $\Bbb{Q}_3((hk^2)^{1/3})$ depending on $hk^2\bmod 9$. – reuns Dec 8 '19 at 10:16
• Sure, but @Philomeno says that he's not familiar with local methods. – nguyen quang do Dec 8 '19 at 10:21