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I'm stuck again at a commutative algebra question. Would love some help with this completion business...

We have a local ring $R$ and $M$ is a $R$-module with unique assassin/associated prime the maximal ideal $m$ of $R$.

i) prove that $M$ is also naturally a module over the $m$-adic completion $\hat{R}$, and $M$ has the same $R$ and $\hat{R}$-submodules.

ii) if $M$ and $N$ are two modules as above, show that $Hom_{R}(M,N) = Hom_{\hat{R}}(M,N)$.

Best wishes and a happy new year!

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2 Answers 2

up vote 2 down vote accepted

Proposition. Let $R$ be a noetherian ring and $M$ an $R$-module, $M\neq 0$. Then $\operatorname{Ass}(M)=\{\mathfrak m\}$ iff for every $x\in M$ there exists a positive integer $k$ such that $\mathfrak m^kx=0$.

Proof. "$\Rightarrow$" Let $x\in M$, $x\neq 0$. Then $\operatorname{Ann}(x)$ is an ideal of $R$. Let $\mathfrak p$ be a minimal prime ideal over $\operatorname{Ann}(x)$. Then $\mathfrak p\in\operatorname{Ass}(M)$, so $\mathfrak p=\mathfrak m$. This shows that $\operatorname{Ann}(x)$ is an $\mathfrak m$-primary ideal, hence there exists a positive integer $k$ such that $\mathfrak m^k\subseteq\operatorname{Ann}(x)$.

"$\Leftarrow$" Let $\mathfrak p\in\operatorname{Ass}(M)$. The there is $x\in M$, $x\neq 0$, such that $\mathfrak p=\operatorname{Ann}(x)$. On the other side, on knows that there exists $k\ge 1$ such that $\mathfrak m^kx=0$. It follows that $\mathfrak m^k\subseteq \mathfrak p$, hence $\mathfrak p=\mathfrak m$.

Now take $x\in M$ and $a\in\hat R$. One knows that $\mathfrak m^kx=0$ for some $k\ge 1$. Since $R/\mathfrak m^k\simeq \hat R/\hat{\mathfrak m^k}$ and $\hat{\mathfrak m^k}=\mathfrak m^k\hat R$, there exists $\alpha\in R$ such that $a-\alpha\in \mathfrak m^k\hat R$. (Maybe Ted's answer is more illuminating at this point.) Now define $ax=\alpha x$. (This is well defined since $\mathfrak m^kx=0$.) Now both properties are clear.

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Here's a claim from which all these statements would follow: For any Cauchy sequence $r_i \in R$, and $x \in M$, the sequence $r_i x$ is eventually constant. (The reason for believing this claim is that, for example, for the first part of i) we have to take a "limit" of $r_i x$ while remaining in $M$, and it's hard to see how else one could do this with no topology on $M$.)

Can we prove this claim? First, let's try this: Since $m$ is an associated prime of $M$, we know that for some $x \in M$, the annihilator of $x$ is $m$. For such an $x$, we can see that $r_i x$ is eventually constant because the Cauchyness of the $r_i$ implies that for some large $N$, we have $r_i - r_j \in m$ for all $i,j \ge N$, hence $r_i x = r_j x$ for all $i,j \ge N$. Unfortunately, this argument doesn't cover all $x \in M$.

However, for a general $x$, if we can show that ann($x$) contains a power of $m$, then we could still apply the same argument as above. Somehow this should follow from the fact that $m$ is the only associated prime ideal of $M$ (and we may have to assume that $R$ is Noetherian). Edit: See YACP's answer for a proof that ann($x$) contains a power of $m$.

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can you please fill in some details on why your claim /YACP's implies my claims? at least the second –  Dquik Jan 10 '13 at 0:25
    
For part ii): It is enough to show that every $R$-module map $f$ is an $\hat{R}$-module map. So let $r \in \hat{R}$. We must show that $f(rm) = rf(m)$. We know $r$ is the limit of a Cauchy sequence $r_i \in R$. From the construction in my answer for part i), $rm$ = $r_i m$ for all $i \ge N$, and $rf(m)$ = $r_i f(m)$ for all $i \ge N'$. Hence for all $i \ge N,N'$ we have $f(rm) = f(r_i m) = r_i f(m) = r f(m)$. –  Ted Jan 10 '13 at 6:05

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