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 $L/K$?

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Eins Null
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    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
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    Just a remark on terminology: If $\mathcal O_K=\mathbb Z[\alpha]$, then $\mathcal O_K$ is sometimes called *monogenic*. – moonlight Apr 25 '16 at 13:59
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    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.) – Gro-Tsen Apr 30 '16 at 23:12
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    This has now been [asked on MathOverflow](http://mathoverflow.net/questions/238037/does-every-ring-of-integers-sit-inside-a-monogenic-ring-of-integers) – Gro-Tsen May 09 '16 at 19:55
  • @Gro-Tsen and sadly it's not had much luck there either! – Mathmo123 May 10 '16 at 14:01
  • Note a similar result holds for the ordinals. See: https://en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form – Jacob Wakem Jun 08 '16 at 22:29
  • What is a ring of integers? Merely a ring containing only integers? – Jacob Wakem Jun 08 '16 at 22:33
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    I honestly don't understand your remark "In fact, there exist number fields which require an arbitrarily large number of elements to form such a basis.": rings of integers are $\mathbb Z$-modules, so all their $\mathbb Z$-bases have the same number of elements (namely the degree of the corresponding field extension). – Matemáticos Chibchas Mar 03 '18 at 04:52
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    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. – Eins Null Apr 15 '18 at 19:38

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