# Does every ring of integers sit inside a ring of integers that has a power basis?

Given a finite extension of the rationals, $$K$$, we know that $$K=\mathbb{Q}[\alpha]$$ by the primitive element theorem, so every $$x \in K$$ has the form

$$x = a_0 + a_1 \alpha + \cdots + a_n \alpha^n,$$

with $$a_i \in \mathbb{Q}$$.

However, the ring of integers, $$\mathcal{O}_K$$, of $$K$$ need not have a basis over $$\mathbb{Z}$$ which consists of $$1$$ and powers of a single element (a power basis). In fact, there exist number fields which require an arbitrarily large number of elements to form such a basis.

Question: Can every ring of integers $$\mathcal{O}_K$$ that does not have a power basis be extended to a ring of integers $$\mathcal{O}_L$$ which does have a power basis, for some finite $$L/K$$?

• Every $abelian$ number field $K$ sits inside $\mathbb Q(\zeta)$ for some root of unity $\zeta$ and has a ring of integers $\mathbb Z[\zeta]$. So your question is answered in the positive for abelian $K/\mathbb Q$.
– RKD
Apr 22 '16 at 23:57
• Just a remark on terminology: If $\mathcal O_K=\mathbb Z[\alpha]$, then $\mathcal O_K$ is sometimes called monogenic. Apr 25 '16 at 13:59
• I think this question should really be reposted on MathOverflow. Also, before tackling the number field case, it's probably worth thinking about curves over a field: given a base curve $C$ and an affine open set $U$, say that $C_1\to C$ (ramified covering) is "monogenic" for these data when there is $f$ regular on $U_1:=U\times_C C_1$ such that $\mathcal{O}(U_1)=\mathcal{O}(U)[f]$. Is it true that there is always $C_2\to C_1$ that is monogenic? (Maybe this is stupidly true or false, I'm not sure. But it could be easier to think about.) Apr 30 '16 at 23:12
• This has now been asked on MathOverflow May 9 '16 at 19:55
• What I should say is this: The number of generators of a ring of integers, as an algebra, can be 1 (monogenic) or arbitrarily large. Apr 15 '18 at 19:38