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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 $p$.

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Who voted to close? And why? –  plusepsilon.de May 8 '13 at 17:05
Is it obvious that the elementary reformulation is the same, given that one is restricted to $p$ unramified and the other is not? –  Will Sawin May 21 '13 at 2:05
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. –  Will Sawin May 21 '13 at 2:31
In our work we only need this for unramified primes. –  Ramin May 21 '13 at 18:58
Have you tried (probably you have) writing down a binary $n$-ic form and playing the same game as in Theorem 9 to Lemma 13 of Bhargava, Shankar, and Tsimerman's paper arxiv.org/pdf/1005.0672.pdf ? This doesn't "just work", the details are much messier, but since you need something weaker I wonder if it is possible to salvage the result you are looking for? –  Frank Thorne May 21 '13 at 23:38
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