# Does every ring of integers sit inside a monogenic ring of integers?

Given a number field $$K/\mathbf{Q}$$ whose ring of integers $$\mathcal{O}_K$$ is, in general, not of the form $$\mathbf{Z}[\alpha]$$ (not monogenic), does there exist an extension $$L/K$$ which has $$\mathcal{O}_L= \mathbf{Z}[\alpha]$$ (is monogenic)?

This question is imported from math stackexchange.

As noted by Jake: Kronecker-Weber implies this is true when $$K/\mathbf{Q}$$ is abelian.

One reason this could be impossible is a local obstruction. Is it the case that, for some $p$, there are no degree $k$ extensions $L$ where $\mathcal O_L \otimes \mathbb Z_p$ is monogenic over $\mathbb Z_p$?

This can happen for small $k$ but there are no obstructions for $k$ sufficiently large. It is sufficient to show that there exists a degree $k$ unramified extension of $K \otimes \mathbb Q_p$, a product of local fields, such that its ring of integers is monogenic. Any element of the product of local rings which is a generator of each local ring (e.g. a root of unity that generates the residue field times 1 + a root of unity) and has a different minimal polynomial in each residue field does the trick. This will always happen when the residue fields are large enough that the number of different irreducible polynomials of generators is at least the number of local fields appearing in the product. If we take our extension to be the product of the degree $k$ unramified extension of each local field appearing in $K \otimes \mathbb Q_p$, then the number of local fields remains $[K:\mathbb Q]$ but the number of irreducible polynomials grows exponentially in $k$.

However, it may not be so easy to find the such an extension. It is already not obvious that there exists an extension satisfying the weaker condition that $\mathbb Z[\alpha]=\mathcal O_K[\alpha]$. Indeed, note that if $\alpha$ has minimal polynomial $f(x)$ over $\mathcal O_K$, of degree $k$, then the discriminant of $\mathcal O_K[\alpha]$ is $\Delta(f) \Delta_K^k$, while the minimal polynomial of $\alpha$ over $\mathbb Z$ is $N_{K/\mathbb Q}( f(x))$, so the discriminant of $\mathbb Z[\alpha]$ is $\Delta(N_{K/\mathbb Q}( f(x)))$. The ratio of these two discriminants is the square of the index of $\mathbb Z[\alpha]$ in $\mathcal O_K[\alpha]$.

We can parameterize the possible polynomials using $k[K:\mathbb Q]$ integer coordinates, $[K:\mathbb Q]$ for each coefficient of $f$. Both $N_{K/\mathbb Q}( f(x))$ and $\Delta(f) \Delta_K^k$ are polynomials in these coordinates. Because their ratio is always an integer, it must be some polynomial in those coordinates. Because it is always a perfect square, it must be the square of some polynomial $g$ of the coordinates.

This index can only equal one if $g= \pm 1$. This is a Diophantine equation.

We can study this equation to try to get an understanding of it even when $\mathcal O_K$ is already monogenic, which might be a good idea. In particular I suspect it might have only finitely many solutions for each $K$ and $k$. If this is so then it is probably hard to prove a solution exists, even if one always does.

• Indeed index form equations are part of the story.- Once one has done explicit computations in number fields of degrees 11 or more, one begins to realize that intuition gained from lower degrees can be awfully misleading. Cyclotomic fields are far from typical. While I have no answer to OP's question, my hunch is that number fields of large degrees with monogenic rings of integers would be the exception rather than the rule, and would tend not to have any nontrivial subfields. – GNiklasch Nov 25 '16 at 13:07

[edit: initially I thought there is no local obstruction but now I'm not sure about that so I deleted what I wrote on that]

Although I don't have proof, I suspect that most fields $K$ are counter examples to the OP's question. Empirically, most number fields are not monogenic. If you compute the best generator that polredabs in GP/PARI can find, and find that it does not generate $O_K$, as is almost always the case (except for low degree or cyclotomic, etc), then it is very likely that $O_K$ is not monogenic. If $O_K$ is monogenic but polredabs didn't find a generator for $O_K$, then $O_K$ would have a generator that is much better than the one found by polredabs. Now polredabs in GP/PARI is based on LLL techniques, and while one can not prove that LLL finds the best element, it is only a bounded factor worse than optimal. This means that the empirical observation that polredabs usually doesn't find a generator for $O_K$ is good empirical evidence that most of these $O_K$ are not monogenetic. It is hard to see how increasing the degree could help globally, so it seems likely that for most fields $K$ (other than very special cases) there will not be an extension $L$ for which $O_L$ is monogenic.