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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 $L/K$?

MeltedStatementRecognizing
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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 $K/\mathbb Q$. – Arkady 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 – 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
  • @Jacob Wakem, if you take a field extension, let's say $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$, then the ring of integers (of $\mathbb{Q}(\sqrt{2})$) refers in some sense to the 'analogon' of what $\mathbb{Z}$ is to $\mathbb{Q}$. In this case the ring of integers would be $\mathbb{Z}[\sqrt{2}]$. But I hope my comment is not misleading, because if $\mathbb{Q}(\alpha)/\mathbb{Q}$ is a field extension, then the ring of integers of $\mathbb{Q}(\alpha)$ is not necessarily(!) $\mathbb{Z}[\alpha]$. – Maty Mangoo Sep 23 '22 at 21:58
  • The difficulty in finding a general approach to construct such an extension arises from the fact that the structure of the ring of integers can be quite complicated, and power bases may not always exist for every number field. – al-Hwarizmi Apr 13 '23 at 06:57

1 Answers1

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the ring of integers of the quadratic field Q(sqrt(2)) has a power basis {1, sqrt(2)}. However, not every ring of integers has a power basis.

consider the ring of integers of a number field that has no real embeddings. For example, let K be the number field Q(sqrt(-5)), which has degree 2 over Q and no real embeddings. The ring of integers of K is Z[(1+sqrt(-5))/2], which has no power basis.

suppose that {a,b} is a power basis for Z[(1+sqrt(-5))/2]. Then every element of Z[(1+sqrt(-5))/2] can be written as a polynomial in a and b with integer coefficients. Since a and b are algebraic integers, their conjugates are also algebraic integers. But the Galois group of K over Q acts transitively on the roots of the minimal polynomial of a and b, so every element of Z[(1+sqrt(-5))/2] has the same set of conjugates. Therefore, the only way to write an element of Z[(1+sqrt(-5))/2] as a polynomial in a and b with integer coefficients is to use the same coefficients for every conjugate, which is not always possible.

the ring of integers of K has no power basis, and therefore cannot be embedded into a ring of integers with a power basis.

Cap_Kidd
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    Looks like more chatGPT nonsense... – user1090793 Apr 21 '23 at 02:46
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    I see my error , not all rings of integers have a power basis, sorry. – Cap_Kidd Apr 21 '23 at 02:52
  • Also you can think I used chatGPT, but I didn’t. Frankly I didn’t even know it could solve equations – Cap_Kidd Apr 21 '23 at 02:52
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  • The ring of integers of $\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Z}[\sqrt{-5}]$ since $-5 \equiv 3 \pmod{4}$. And ${1, \sqrt{-5}}$ is a power basis for $\mathbb{Z}[\sqrt{-5}]$. 2) You haven't attempted to answer the question. The OP clearly states that not every ring of integers has a power basis. The question is if there is a further extension of $K$ whose ring of integers has a power basis. Your last sentence claims that this is impossible with no proof.
    • – Viktor Vaughn Apr 21 '23 at 03:05
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    I see that. I haven’t answered the question correctly and wasted your time, sorry. – Cap_Kidd Apr 21 '23 at 03:07