146
$\begingroup$

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 $L/K$?

$\endgroup$
9
  • 18
    $\begingroup$ 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 $K/\mathbb Q$. $\endgroup$
    – RKD
    Apr 22 '16 at 23:57
  • 9
    $\begingroup$ Just a remark on terminology: If $\mathcal O_K=\mathbb Z[\alpha]$, then $\mathcal O_K$ is sometimes called monogenic. $\endgroup$
    – moonlight
    Apr 25 '16 at 13:59
  • 2
    $\begingroup$ 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.) $\endgroup$
    – Gro-Tsen
    Apr 30 '16 at 23:12
  • 3
    $\begingroup$ This has now been asked on MathOverflow $\endgroup$
    – Gro-Tsen
    May 9 '16 at 19:55
  • 1
    $\begingroup$ 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. $\endgroup$
    – Eins Null
    Apr 15 '18 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.