In this post we will say something about the completion of adic rings.
Suppose is a topological ring, and is an ideal of definition for , then we set . We call the Hausdorff completion of .
Note that the original definition for Hausdorff completion is to take the topological completion using Cauchy sequences and then module the closure of the zero element. We can show that for a topological ring, these two definitions are the same. Thus is a Hausdorff space(this can also be seen from the first definition, since each is a discret space). Moreover, this shows that the above definition is independent of the choice of the ideal of definition.
As is expected, the Hausdorff completion is characterized by some universal property. That is:
is a topological ring, and its Hausdorff completion with the morphism of topological rings , then for any Hausdorff complete topological ring and any morphism of topological rings , there exists a unique morphism of topological rings such that .
Moreover, we can see that .
As for adic rings, we have that:
If is an adic ring, then is also an adic ring. Suppose is a ring of definition for and is an ideal of definition associated to for , then is a ring of definition for , and is an ideal of definition. Besides, there is an isomorphism of topological rings between and .
It might be interesting to see what the morphism between these two rings is. For any element (viewed as a Cauchy sequence in ), then according to the definition of Cauchy sequence, there is an such that for any we have that since is open. In this way, the element actually lies in (we disregard of the first finite terms), so we get a decomposition with the first factor in and the second factor in . So we can define . It is not hard to show that is well defined, that is to say, if there is another representative equivalent to , and another decomposition of , then there image are the same. This is a simple verification.