Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello, Let $K/{\mathbb Q}$ be a finite extension which is not necessarily Galois, and ${\mathcal O}$ be the ring of integers of $K$. Let $p$ be a prime in ${\mathbb Q}$ and let $p {\mathcal O}={\mathfrak p}_1^{k_1} \cdots {\mathfrak p}_r^{k_r}$ be its prime factorization in $K$. Next let ${\mathcal O}_i$ be the ring of integers of the completion of $K$ at ${\mathfrak p}_i$. Let ${\mathcal O}_p={\mathcal O}_1^{k_1} \times \cdots \times {\mathcal O}_r^{k_r}$. Is there a description of subrings $R$ of ${\mathcal O}_p$ such that $[{\mathcal O}_p:R ] < \infty$ (additive subgroup index) ? Is it for example true that "most" of such $R$'s are of the form $R_1 \times \cdots \times R_r$? Here "most" should mean the following: Let ${\mathcal n}_t$ be the number of subrings $R$ such that $[{\mathcal O}_p: R] \leq t$ and let ${\mathcal n}'_t$ be the number of subrings $R$ of index at most $t$ which are expressible as a direct product $R_1 \times \cdots \times R_r$. Then is it true that $n'_t / n_t \to 1$ as $t \to \infty$? It would of course be most desirable if $n_t = n'_t$. My sincerest apologies ahead of time if this turns out to be a stupid question.

Added in revision: Thank you for the answer and the comment. I would like to think of Laurent's example as ${\mathbb Z}_p^2$ and not ${\mathbb Z}_p \times {\mathbb Z}_p$. So it seems that I posed the question incorrectly; what I really need is to group together those ${\mathcal O}_i$'s that are isomorphic to each other, even if the the corresponding ${\mathfrak p}_i$'s are different.

A word about the genesis of this problem: Nathan Kaplan and I have recently proved a theorem about counting the number of subrings of ${\mathbb Z}^n$ for small $n$ using $p$-adic integration techniques (we treat $n \leq 5$, the case $n=5$ seems to be new). Now I'm trying to see if we can use what we have proved to count orders in quintic fields (for cubic fields this is due to Davenport and Heilbronn, and for quartic fields this is a consequence of Nakagawa's work). To illustrate the idea, for a moment imagine that ${\mathbb Q}$ has a quintic Galois extension $K$. Except for finitely many exceptions, a given prime of $p$ either remains prime or splits as a product ${\mathfrak p}_1 \cdots {\mathfrak p}_5$. The situation for split primes is the same as counting the number of subrings of ${\mathbb Z}_p^5$ which we know how to do. The inert primes require separate treatment.

It turns out that the question is quite a bit more interesting than I had originally thought (what else is new!).

share|cite|improve this question
Ramin -- do you expect there to be a "mass formula" for orders, as one sees in Serre, Kedlaya, Bhargava, Matchett Wood's work? (see e.g. Kedlaya's paper –  JSE Jan 22 '11 at 3:59
@Esperantist: Thank you! This is awesome. As of yesterday I suspected that most subrings should be non-split, but I wasn't able to construct an example. This is very nice. –  Ramin Jan 23 '11 at 4:56
@Jordan: The p-adic integration idea that Nathan and I use would count orders on the nose (a copy is available on my homepage). I don't know; maybe obtaining a mass formula is the right thing to do, but I don't know how to do that using our methods. AV and I were at some point trying to count (some particular) orders using the trace formula. We never completed the project, but in that case we would have obtained a mass formula involving regulators and sizes of automorphism groups. –  Ramin Jan 23 '11 at 5:05
@Esperantist: Comment deleted and identity protected! –  Ramin Jan 24 '11 at 2:26

1 Answer 1

up vote 3 down vote accepted

Assume $\mathcal{O}_p=\mathbb{Z}_p\times\mathbb{Z}_p$.

The only subring of finite index in $\mathbb{Z}_p$ is $\mathbb{Z}_p$: indeed such a ring must contain $n\mathbb{Z}_p$ for some $n$, hence $p^k\mathbb{Z}_p$ for some $k$, hence corresponds to a subring of $\mathbb{Z}/p^k\mathbb{Z}$. (Of course I take it that your subrings contain 1).

Hence $n'_t=1$ for all $t$, while $n_t>1$ as soon as $t\geq p$: consider $\{(x,y)\in\mathbb{Z}_p\times\mathbb{Z}_p\mid x\equiv y\pmod p\}$.

share|cite|improve this answer
Nice example ! And condering congruences with various powers of $p$ shows that the $n'_t/n_t$ tends to $0$ ! –  Qing Liu Jan 19 '11 at 14:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.