Huber spaces-5

In this post we will say something about the completion of f-adic rings.

Suppose A is a topological ring, and I_0 is an ideal of definition for A, then we set \hat{A}=\varprojlim A/I_0^n. We call \hat{A} the Hausdorff completion of A.

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 \hat{A} is a Hausdorff space(this can also be seen from the first definition, since each A/I_0^n 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:

A is a topological ring, and \hat{A} its Hausdorff completion with the morphism of topological rings i:A\rightarrow \hat{A}, then for any B Hausdorff complete topological ring and any morphism of topological rings f:A\rightarrow B, there exists a unique morphism of topological rings f':\hat{A}\rightarrow B such that f'\circ i=f.

Moreover, we can see that ker(i)=\bar{\{0\}}.

As for f-adic rings, we have that:

If A is an f-adic ring, then \hat{A} is also an f-adic ring. Suppose B is a ring of definition for A and I is an ideal of definition associated to B for A, then \hat{B}=\varprojlim B/I^n is a ring of definition for \hat{A}, and \hat{I}=\varprojlim I/I^n is an ideal of definition. Besides, there is an isomorphism of topological rings between A\bigotimes_B\hat{B} and \hat{A}.

It might be interesting to see what the morphism between these two rings is. For any element (a_n)_n\in \hat{A}(viewed as a Cauchy sequence in A), then according to the definition of Cauchy sequence, there is an N such that for any n\geq N we have that a_n-a_N\in B since B is open. In this way, the element (a_n-a_N)_n actually lies in \hat{B}(we disregard of the first finite terms), so we get a decomposition (a_n)_n=(a_n-a_N)_n+(a_N)_n with the first factor in \hat{B} and the second factor in A. So we can define j: \hat{A}\rightarrow \hat{B}\bigotimes_BA, (a_n)_n\mapsto (a_n-a_N)_n\bigotimes_B1+1\bigotimes_Ba_N. It is not hard to show that j is well defined, that is to say, if there is another representative (a_n')_n equivalent to (a_n)_n, and another decomposition of (a_n')_n, then there image are the same. This is a simple verification.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s