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Suppose $A$ is a commutative noetherian ring, $\mathfrak{a} \subseteq A$ is an ideal, and $M,N$ are two $\mathfrak{a}$-adically complete $A$-modules (complete = complete and separated). Is the $A$-module $\operatorname{Hom}_A(M,N)$ also an $\mathfrak{a}$-adically complete module? If it is not true in general, what if we assume that $A$ is complete?

Thank you for any help!

Edit: Here are some of my thoughts.

First, if $A$ is complete and $M$ and $N$ are finitely generated, then the claim is true, as finitely generated modules over a complete ring are complete.

Second, it is actually equivalent to assume that only $N$ is complete. This is because we have an isomorphism $\operatorname{Hom}_A(\Lambda_I M, \Lambda_I N) \cong \operatorname{Hom}_A(M,\Lambda_I N)$.

Thus, it follows that I am actually asking if the hom into a complete module is always complete.

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up vote 3 down vote accepted
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I believe that the answer is "yes".
I will not assume that $A$ is Noetherian in what follows, but will make more specific assumptions as necessary, which will hopefully add clarity.


Some preliminary remarks:

  1. One remark is that I don't think it is any loss of generality to assume that $A$ is $\mathfrak a$-adically complete. The reason is that (writing $\hat{A}$ to denote the $\mathfrak a$-adic completion of $A$), if $M$ and $N$ are $\mathfrak a$-adically complete, then the $A$-action on $M$ and $N$ extends canonically to an $\hat{A}$-action, and any $A$-linear homomorphism between them is necessarily $\hat{A}$-linear.

  2. If $N$ is $\mathfrak a$-adically separated, then it is clear that the same is true of $Hom_A(M,N)$ (since if $\varphi \in \mathfrak a^i Hom_A(M,N)$, then $\varphi(M) \subset \mathfrak a^i N$).

  3. If $M$ is finitely presented (equivalently, finitely generated when $A$ is Noetherian), then applying $Hom_A(M,\text{--})$ to a finite presentation $A^m \to A^n \to M \to 0$ of $M$, we obtain an exact sequence $$0 \to Hom_A(M,N) \to N^n \to N^m.$$ Thus in this case $Hom_A(M,N)$ is $\mathfrak a$-adically complete, since it is the kernel of a map between $\mathfrak a$-adically complete modules.


The main part:

Let $\varphi_i \in \mathfrak a^i Hom_A(M,N)$, for $i \geq 0$. If $m \in M$, then we find that $\varphi_i(m) \in \mathfrak a^i N,$ and hence that $\varphi(m):= \sum_{i = 0}^{\infty} \varphi_i(m)$ is well-defined in $N$ (since $N$ is $\mathfrak a$-adically complete). The map $m \mapsto \varphi(m)$ gives a well-defined homomorphism from $M$ to $N$.

One would like to say that $\varphi$ is the $\mathfrak a$-adic sum of the $\varphi_i$ in $Hom_A(M,N),$ and hence that $Hom_A(M,N)$ is $\mathfrak a$-adically complete. However, I don't see how to do this in general. The problem is that for this, one needs to show that $\sum_{i \geq i_0} \varphi_{i}$ (defined pointwise, as in the preceding construction) actually lies in $\mathfrak a^{i_0} Hom_A(M,N)$ (or at least in $\mathfrak a^j Hom_A(M,N)$ for some value of $j$ which goes to $\infty$ as $i_0$ does), and I don't see how to do this in general. (Each map $\varphi_i$ for $i \geq i_0$ is in $\mathfrak a^{i_0} Hom_A(M,N)$ by assumption; the problem is to deduce that $\sum_{i \geq i_0} \varphi_{i}$, which is defined pointwise, is also in there. This involves effecting some kind of "division by $\mathfrak a^{i_0}$" in $Hom_A(M,N)$, which I don't see how to do in general.)

Suppose now that $\mathfrak a$ is finitely generated (which certainly holds if $A$ is Noetherian). Then $\mathfrak a^{i_0}$ is finitely generated, say by elements $a_1,\ldots,a_s$, and so for each $i \geq i_0$, we may write $\varphi_i = \sum_{j = 1}^s a_j \psi_{i,j}$, where $\psi_{i,j} \in \mathfrak a^{i-i_0} Hom_A(M,N)$. (Here I am using the fact that $\varphi_{i} \in \mathfrak a^i Hom_A(M,N) = \sum_{j = 1}^s a_j \mathfrak a^{i-i_0} Hom_A(M,N)$.)

Now we may define $\psi_j := \sum_{i \geq i_0} \psi_{i,j}$ pointwise as an element of $Hom_A(M,N)$, for $j = 1,\ldots,s$ (again using the fact that $N$ is $\mathfrak a$-adically complete). On the one hand, one easily checks (just by working pointwise) that $\sum_{j=1}^s a_j \psi_j = \sum_{i \geq i_0} \varphi_i$. On the other hand, one sees that$\sum_{j = 1}^s a_j \psi_j \in \mathfrak a^{i_0} Hom_A(M,N).$ We conclude that indeed $\sum_{i \geq i_0} \varphi_i$ does lie in $\mathfrak a^{i_0} Hom_A(M,N),$ and hence that $\varphi$ is the $\mathfrak a$-adic sum of the $\varphi_i$.

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