If $A$ is a noetherian local ring and $M$ an $A$-module, then we define the completion $\hat{M}$ of $M$ with respect to the stable $\mathfrak{m}$-filtration $\{M_n\}$ by $$\hat{M}=\left\{(a_1,a_2,...)\in\prod_{i=1}^\infty M/M_i:a_j\equiv a_i\bmod{M_i}\,\,\forall j>i\right\}.$$ See also my previous question.

Now in the book I use (A SINGULAR Introduction to Commutative Algebra by Greuel/Pfister), there is no universal property of this completion mentioned, but once it uses something that looks like one: We have a map from $K[x_1,...,x_n]_{\langle x_1,...,x_n\rangle}$ to some complete ring, hence we got a map from $K[[x_1,...,x_n]]$ to it. Is that the 'universal property of the completion of a ring / module', and if yes, is it somehow obvious from my definition of the completion, so that we could use it directly?

How to prove this property with the above definition? (If it really works for modules; I don't know, at least for rings I guess it should be something like: If $A\to B$ is a ring homomorphism and $B$ is complete, then there is a unique map $\hat{A}\to B$ extending it). Well, I think I maybe got a clue right now, and you could perhaps tell me if this is the correct way (I'd still like to know if this is 'the' universal property of the completion):

If $A\to B$ is a ring homomorphism ($A$ and $B$ noetherian local rings; does this homomorphism have to be local, too?), and $B$ is complete, then I get and induced map $\hat{A}\to\hat{B}=B$ as wanted.

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    $\begingroup$ To define a homomorphism from $K[[x_1,\ldots ,x_n]]\to B$ a complete local ring, it suffices to give the images of $x_i$ such that the homomorphism is local, i.e. the maximal ideal maps into the maximal ideal. Thus the map $K[x_1,\ldots ,x_n]_{(x_1,\ldots ,x_n)}\to B$ uniquely defines such a homomorphism. $\endgroup$
    – Andrew
    Jul 11, 2012 at 19:25
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    $\begingroup$ @Andrew: Oh, I was a little misguided by the text in the book there it seems (it's formulated roughly as 'since $A$ is complete, we can extend ...'). The way you put it sounds much more natural to me, thanks! I don't even have to use that $K[[x_1,...,x_n]]$ is the completion of the localized ring. $\endgroup$ Jul 11, 2012 at 19:36

1 Answer 1


As you are considering modules endowed with stable $m$-filtrations, these are simply $A$-modules with $m$-adic topology (defined by the filtration $(m^nM)_n$).

The completion $\hat{M}$ satisfies the following universal property in the category of $m$-adic $A$-modules:

  1. There exists an $A$-linear map $i: M\to \hat{M}$,

  2. For any complete $A$-module $N$ for any $A$-linear map $f : M\to N$, there exists a unique factorization of $f$ as $i : M\to \hat{M}$ and $\hat{f} : \hat{M} \to N$.

Proof: (1) the map $i$ is given by $i(x)=(x \mod mM, x \mod m^2M, ...)$. (2) The map $f$ induces $M/m^nM \to N/m^nN$. Passing to the limit we get a map $\hat{f} : \hat{M}\to \hat{N}=N$. As $i(M)$ is dense in $\hat{M}$, for any factorization as in (2), $\hat{f}$ is uniquely determined by $f$.

For homomorphisms of local rings $\rho : A\to B$ you have to require the maps to be local to insure continuity, then $\rho$ induces $$ A/m_A^n \to B/m_B^n $$ for all $n\ge 1$. Passing again to the limit we get a ring homomorphism $\hat{A}\to \hat{B}$ which is uniquely determined by $\rho$ because $A$ is dense in $\hat{A}$.


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