Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have checked the list of similar titles, proposed by the site. I hope this is not a repetition.
This question arises from a proof of a proposition in the book Basic Number Theory, as follows.

Every finitely generated $R$-module $M$ is the direct sum of finitely many summands, each of which is isomorphic either to R or to a module R/$P^v$, $v$>0.
Let M be generated by elements $m_1,...,m_n$. Take a vector-space $v$ of dimension n over $K$, with a basis $v_1,...v_n$; put $L$=$\Sigma$ $Rv_i$. Then the formula
$\Sigma$ $x_iv_i$=$\Sigma$ $x_im_i$,
where the $x_i$ are taken in $R$ for i from 1 to n, defines a morphism of $L$ onto $M$; therefore $M$ is isomorphic to $L/M'$, where $M'$ is the kernel of that morphism. Apply now corollary 1 to $L$ and $M$; as $M$ is a subset of $L$, we have $v_j$is positive for j from 1 to r, and r=s.

After that proof, Weil then defines $n_i$ as the dimension of $R_i=M_i/M_{i+1}$, where $M_i$:=$\pi^iM$, over the residual field $k=R/P$; moreover, after the proposition is set up, one can define the number$N_v$ as $N_0$=the number of times R appears in the summands, and $N_v$=the number of times R/$P^v$ appears in the summands. Now, Weil asserts that $n_i$=$N_0$+$\Sigma_{v>i}N_v$. And my question is why is this true.
Here, R is the maximal compact subring of a p-field $k$, in which $P$ is the maximal ideal, where a $p$-field is a non-discrete locally compact field, either of characteristic $p$, or a finite extension of $Q_p$ the completion of $Q$ with respect to the $p$-adic valuation.
I may not exprese my question explicitly enough, so feel free to ask for explanations, and I will do my best to clarify things.
Thanks and regards here.

share|cite|improve this question
Hi. Just as a note, the kernel "M" is different from the module "M". And $R_i=M_i/M_{i+1}$. – John M Jul 25 '11 at 15:06
Note: this is on pg. 30 of Weil's book. I'll try to have a look at your question later, but I'm sure that John's answer is great. – Dylan Moreland Jul 25 '11 at 16:25
@Dylan Moreland - don't assume it's great! My first answer was bogus. – John M Jul 25 '11 at 16:30
Here's a question to anyone who has Weil's book: what are the actual conditions on the ring $R$ here? Does $R$ merely have to be Noetherian, and $P$ can be any maximal ideal? – John M Jul 25 '11 at 16:42
Also, Weil's proposition is also a corollary of the usual structure theorem for finite modules over a principal ring, isn't it? – Dylan Moreland Jul 26 '11 at 3:27
up vote 2 down vote accepted

Here's my stab at this: Here I assume $R$ is a DVR with max ideal $P=(\pi)$. If this isn't correct, let me know.

If the module $M \cong R^{N_0} \oplus (R/P)^{N_1}\oplus\dots$, then $M_i=\pi^iM \cong(P^i)^{N_0}\oplus(P^i/P^{i+1})^{N_{i+1}}\oplus(P^i/P^{i+2})^{N_{i+2}}\oplus\dots$, since the rest of the terms with index less than or equal to $i$ are trivial.

So then we have $M_i/M_{i+1}\cong (P^i/P^{i+1})^{N_0}\oplus(P^i/P^{i+1})^{N_{i+1}}\oplus(P^i/P^{i+1})^{N_{i+2}}\oplus\dots$.

Note that $\text{dim}_{R/P}(P^i/P^{i+1})=1$, so $$n_i=\text{dim}_{R/P}(M_i/M_{i+1})=N_0+\sum_{\nu>i}N_i$$

share|cite|improve this answer
Looks good to me! – Dylan Moreland Jul 26 '11 at 4:03
In fact, one of the main problems I encountered before is that, although I decomposed the ring, I do not know why we can view as trivial those terms with index less than $i$; this is suddenly solved. Thank you. – awllower Jul 26 '11 at 11:21

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.