Topologies determine the absolute value October 5, 2009Posted by Akhil Mathew in algebraic number theory, algebra, number theory.
Tags: absolute values, topologies, norms, completeness
Time to go back to basic algebraic number theory (which we’ll need for two of my future aims here: class field theory and modular representation theory), and to throw in a few more facts about absolute values and completions—as we’ll see, extensions in the complete case are always unique, so this simplifies dealing with things like ramification. Since ramification isn’t affected by completion, we can often reduce to the complete case.
Henceforth, all absolute values are nontrivial—we don’t really care about the absolute value that takes the value one everywhere except at zero.
I mentioned a while back that absolute values on fields determine a topology. As it turns out, there is essentially a converse.
Theorem 1 Let , be absolute values on inducing the same topology. Then is a power of .
So, first of all, given , the sequence converges to zero iff , and iff . Thus we see that the unit disks are the same under both absolute values.
we have to show
as we can then use symmetry to show that , and then one gets that , differ by the same power as the ‘s.
First, I claim this is true for . Then (1) says
Next, I claim that in the complete case, absolute values are very often going to be equivalent.
So fix a field complete with respect to the absolute value . Let be a finite-dimensional vector space over . A norm on isa map satisfying all the usual constraints, i.e. scalar multiplicativity (with respect to ) and the triangle inequality.
Theorem 2 All norms on induce the same topology.
For the proof, I’ll make the simplifying (but unnecessary) assumption that is locally compact—this is the case for the real and complex numbers, finite extensions of the -adic fields, and finite extensions of power series fields (all of which go under the name “local fields”).
Fix a basis . We can choose the norm on : if , then . Then from the triangle inequality it is easy to check that for all , for some large . If we show the reverse, that there exists a with for all , then the assertion will be complete. By homogeneity we may assume ; the set of such forms a compact subset of (under the -topology!), on which by the previous inequality is a continuous and nonzero function. It has a minimum, which is the referred to.
We can apply this to the case of fields:
Corollary 3 Given a complete field and a finite extension , the absolute value on can be extended to in precisely one way.
The above result implies uniqueness; we’ve shown existence earlier, at least in the discrete case. So in particular if is a complete DVR and the integral closure in some extension of the quotient field of of degree , and if prolongs , then the basic ramification formula becomes
Things are thus a lot simpler.