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Let $K$ be a degree $n$ extension of ${\mathbb Q}$ with ring of integers $R$. An order in $K$ is a subring with identity of $R$ which is a ${\mathbb Z}$-module of rank $n$.

Question: Let $p$ be an unramified prime in $K$. Is it true that the number of orders in $R$ of index equal to $p^r$, for some natural number $r$, is less than or equal to the number of subrings with identity of ${\mathbb Z}^n$ of index equal to $p^r$?

Nathan Kaplan and I need this fact for $n=5$ in a project where we are trying to find asymptotic formulae for the number of orders of bounded index in a given quintic field.

We've been staring at Jos Brakenhoff's thesis for a while, but I haven't gotten anywhere. Any advice will be greatly appreciated. Thanks.

Added in edit: Here is an elementary reformulation of this problem. Let $r(x)$ be a polynomial with integer coefficients. Then show that for any natural number $a$ in order to maximize the number of subrings of $({\mathbb Z}/p^a {\mathbb Z})[x]/(r(x))$ of a given index, the polynomial $r(x)$ has to be a product of linear factors modulo $p$.

Update: In a soon-to-be-posted preprint Nathan Kaplan, Jake Marcinek, and I have proved an asymptotic formula for the number of orders in a given quintic field of bounded discriminant, using an argument independent of the above question. We can also give non-trivial upper bounds for a general number field. I'm now convinced the answer to the question is yes, but I still don't know how to prove it.

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    $\begingroup$ Who voted to close? And why? $\endgroup$
    – Marc Palm
    Commented May 8, 2013 at 17:05
  • $\begingroup$ Is it obvious that the elementary reformulation is the same, given that one is restricted to $p$ unramified and the other is not? $\endgroup$
    – Will Sawin
    Commented May 21, 2013 at 2:05
  • $\begingroup$ It is clear that this question can be studied locally at $p$. In other words, this depends only on $R \otimes \mathbb Z_p$, so it depends only on $K \otimes \mathbb Q_p$, which is just a product of $p$-adic fields. $\endgroup$
    – Will Sawin
    Commented May 21, 2013 at 2:31
  • $\begingroup$ In our work we only need this for unramified primes. $\endgroup$
    – Ramin
    Commented May 21, 2013 at 18:58
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    $\begingroup$ I found a paper by Melanie Wood which might be relevant, although I did not check it carefully: arxiv.org/pdf/1007.5508v1.pdf $\endgroup$
    – Frank Thorne
    Commented May 21, 2013 at 23:41

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