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 think I came up with the following result. But I'm not 100% sure. Is this correct? If yes, how does one prove this?

Theorem? Let $A$ be a discrete valuation ring, $K$ its field of fractions. Let $L$ be a finite separable extension of $K$. Let $B$ be the integral closure of $A$ in $L$. Let P be the maximal ideal of A. Let $Q_i$, $i$ = $1, ..., r$ be the maximal ideals of $B$ lying over $P$. Let $M$ be a finitely generated torsion-free module over $B$. Let $\hat{M_P}$ be the completion of $M$ with respect to $P$-adic topology. Let $\hat{M_{Q_i}}$ be the completion of $M$ with respect to $(Q_i)$-adic topology. Then $\hat{M_P}$ $\cong$ $\prod_{i}\hat{M_{Q_i}}$

EDIT I need this to prove this theorem.

share|cite|improve this question
As $M$ is free and finite over $B$, the proof is same as in – user18119 May 20 '12 at 6:51
@Qil Right! I forgot B is a principal ideal domain. :-) Thanks. – Makoto Kato May 20 '12 at 7:22
Your "theorem" holds for any noetherian local ring $A$ and any finitely generated $B$-module $M$. The proof is contained in your answer in the above question. – user18119 May 20 '12 at 21:28
@Qil That's interesting. I guess I need some effort to prove it, though. – Makoto Kato May 22 '12 at 21:37
up vote 3 down vote accepted

For any $n\ge 1$, $B/P^nB$ is Artinian with maximal ideals $Q_i/P^nB$, so the canonical map $$ B/P^nB \to \prod_{1\le i\le r} B_{Q_i}/P^nB_{Q_i}$$ is an isomorphism. Tensoring by $M$ over $B$ we get a canonical isomorphism $$ M/P^n M\simeq \prod_{1\le i\le r} M_{Q_i}/P^nM_{Q_i}.$$ As $Q_i^NB_{Q_i}\subseteq PB_{Q_i}\subseteq Q_iB_{Q_i}$ for some $N\ge 1$, we get $$ \widehat{M}=\varprojlim_n M/P^nM\simeq \prod_{1\le i\le r} \widehat{M_{Q_i}}.$$ In fact the theorem you cited is not used here (it is useful if we take inverse limit of the first isomorphism before tensoring by $M$).

share|cite|improve this answer
Thanks. I need to digest your proof. Please wait for a while until I accept it. – Makoto Kato May 23 '12 at 0:24

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.