# $\hat{M_P}$ $\cong$ $\prod_{i}\hat{M_{Q_i}}$

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.

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As $M$ is free and finite over $B$, the proof is same as in math.stackexchange.com/questions/137888. – 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

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 $M$).