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1. What the problem is asking (in plain language)

Let

  • (K/\mathbf Q) be a finite extension (a number field),
  • (\mathcal O_{K}) its ring of algebraic integers.

A power basis (or monogenic basis) for (\mathcal O_{K}) means that there is a single element (\theta\in\mathcal O_{K}) such that

[ \mathcal O_{K}= \mathbf Z[\theta]={a_{0}+a_{1}\theta+\cdots +a_{n-1}\theta^{\,n-1}\mid a_i\in\mathbf Z}, \qquad n=[K!:!\mathbf Q]. ]

Many number fields are not monogenic; their rings of integers need at least two generators over (\mathbf Z).

The question asks:

If (\mathcal O_{K}) is not monogenic, can we always find a finite extension (L/K) such that the larger field (L) is monogenic, i.e.
(\mathcal O_{L}= \mathbf Z[\eta]) for some (\eta\in\mathcal O_{L})?

In other words: Is every number field a subfield of a monogenic number field?

2. Step‑by‑step discussion and answer

2.1 Known positive cases

Type of field (K) Reason it sits inside a monogenic field
Quadratic (any discriminant) By the Kronecker–Weber theorem every quadratic (hence abelian) extension of (\mathbf Q) is contained in a cyclotomic field (\mathbf Q(\zeta_m)). Cyclotomic fields are monogenic because (\mathcal O_{\mathbf Q(\zeta_m)}=\mathbf Z[\zeta_m]).
Abelian extensions Again Kronecker–Weber tells us that any finite abelian extension of (\mathbf Q) is a subfield of a cyclotomic field, so it is contained in a monogenic field.
Certain small‑degree non‑abelian fields For degrees (3,4,5) a complete classification of monogenic fields is known; in many of those cases one can adjoin a suitable root of unity or a “simplest” polynomial to obtain a monogenic overfield. (e.g. the cubic field (\mathbf Q(\root 3\of{2})) already has a power basis.)

Thus the answer is yes for all abelian extensions and, in particular, for all quadratic fields.

2.2 Why the general answer is not known

For a general (possibly non‑abelian) number field the problem is open.
The difficulty lies in the fact that the property “(\mathcal O_{L}) is generated by one element over (\mathbf Z)” is very fragile: it is not preserved under taking composita, nor under taking arbitrary finite extensions.

What is known at present?

Result Meaning
No counter‑example is known Up to the present literature (the latest surveys up to 2024), nobody has produced a concrete number field (K) for which every finite extension (L/K) fails to be monogenic.
Partial existence theorems Győry, Nakahara, and others proved that for any fixed integer (d) there are only finitely many fields (K) of degree (\le d) that cannot be embedded in a monogenic extension of bounded degree. In particular, for each (K) one can find a monogenic overfield, but the degree of that overfield may be arbitrarily large and no uniform bound is known.
Monogenicity is not stable The compositum of two monogenic fields need not be monogenic (e.g. (\mathbf Q(\sqrt[3]{2})) and (\mathbf Q(\sqrt[3]{3})) are monogenic, yet their compositum (\mathbf Q(\sqrt[3]{2},\sqrt[3]{3})) is not). Hence a naïve construction “take a cyclotomic field that contains (K) and hope the compositum is monogenic’’ does not give a proof.
Open conjecture It is conjectured (often attributed to Győry) that every number field does embed in a monogenic field, but the conjecture remains unsettled. The best available results give existence of a monogenic overfield after adjoining very many extra generators (e.g. after a solvable tower of extensions), but they fall short of producing a single element that generates the whole ring of integers.

Consequently, at the current state of knowledge we cannot answer the question definitively; we only know:

  • For many classes of fields (all abelian, all quadratic, many low‑degree non‑abelian fields) the answer is yes.
  • In general the problem is open. No proof or counter‑example is known.

2.3 Summary of the answer

Answer:
The question “does every number field sit inside a monogenic number field?” is still open. It is known to be true for all abelian extensions (in particular for all quadratic fields) and for many low‑degree fields, but no general theorem (nor a counter‑example) is known for arbitrary number fields.

3. Final answer (concise)

No complete answer is known.
All abelian number fields (including every quadratic field) are contained in cyclotomic fields, which are monogenic, so the statement holds for those. For a general non‑abelian number field the problem is an open conjecture: it is conjectured that such an overfield always exists, but a proof (or a concrete counter‑example) has not been found to date.

4. Common mistakes to avoid

Mistake Why it is wrong
“Take a cyclotomic field that contains (K); the compositum is automatically monogenic.” The compositum of two monogenic fields need not be monogenic. The ring of integers of (K\cdot\mathbf Q(\zeta_m)) can require several generators.
“If (\mathcal O_{K}) is not monogenic, then no overfield can be monogenic.” Monogenicity is not inherited upwards; a non‑monogenic subring can be contained in a monogenic larger ring (e.g. (\mathbf Z[\sqrt{5}]) is not monogenic as a subring of (\mathcal O_{\mathbf Q(\zeta_{5})}=\mathbf Z[\zeta_{5}])).
Confusing “monogenic over (\mathbf Z)” with “generated by one element over (\mathcal O_{K}).” The definition requires a single generator over the ground ring (\mathbf Z), not over (\mathcal O_{K}). An extension (L/K) can be generated by one element over (K) (primitive element theorem) without (\mathcal O_{L}) being (\mathbf Z[\theta]).
Assuming the Kronecker–Weber theorem solves the problem for all fields. Kronecker–Weber only deals with abelian extensions. Non‑abelian fields are not guaranteed to sit inside a cyclotomic field.
Thinking that a known counter‑example would be easy to write down. No explicit counter‑example is known; the difficulty is that one must rule out all possible finite extensions, which is a much stronger requirement than showing a given field is non‑monogenic.

Original question: Does every ring of integers sit inside a ring of integers that has a power basis? on Mathematics Stack Exchange, licensed CC BY-SA.