trivial valuation
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trivial valuation
The trivial valuation of a field is the Krull valuation of such that and for other elements of .
Properties
1. Every field has the trivial valuation.
2. The trivial valuation is non-archimedean.
3. The valuation ring of the trivial valuation is the whole field and the corresponding maximal ideal is the zero ideal.
4. 5. A finite field has only the trivial valuation. (Let be the primitive element of the multiplicative group of the field, which is cyclic. If is any valuation of the field, then one must have since otherwise . Consequently, for all non-zero elements .)
6. Every algebraic extension of finite fields has only the trivial valuation, but every field of characteristic 0 has non-trivial valuations.
Major Section:
Reference
Type of Math Object:
Definition
Parent:
Mathematics Subject Classification
12J20 General valuation theory11R99 None of the above, but in MSC2010 section 11Rxx
Comments
6.
part 6. isn't correct. the algebraic closure of any finite field has no non-trivial valuation. maybe better to say that a field of characterstic 0 has a non trivial valuation?
Re: 6.
You state a surprising thing of the alg. closure. I must check it. Can you please justify it (or make an entry of it)?
Jussi
Re: 6.
im new to this, so i can say here:
I only learned these things a couple of days ago!) :)
ps if you see any errors please tell me!
the algebraic closure for F_q follows like this:
if G_q is the algebraic closure of F_q, and v is a valuation on G_q,
if it is non trivial on some element, then it is non trivial on some
finite extension of F_q, but all finite fields ( and hence finite
extensions of F_q) have only the trivial valuation. so v is trivial on
G_q.
the more general fact is this:
If you have a (maybe infinite) algebraic extension L/K and a valuation
on L that restricts to the trivial valuation on K, then it must be
trivial on L also. this is proved as follows:
if L/K is a finite algebraic extension, v a valuation on L that is
trivial on K. then take any basis a_i for L/K (as a vector space).
then you have
v(a_1*k_1 + ... + a_n*k_n) <= v(a_1 k_1)+...+v(a_n*k_n)=v(a_1)+...+v(a_n).
But the last sum is just a constant, so the valuation is bounded on L,
which means it can only take the values 0 or 1 (if any other then you
could take powers to get an arbitrarily large valuation).
the infinite case follows since if you have a valuation on an
algebraic extension L/K, that is trivial on K, and has v(j) \neq 0 for
some j\in L, then K(j)/K is a finite algebraic extension so that v
must be trivial on K(j) too, hence v(j)=1, a contradiction.
Re: 6.
You are fully right, Lance, and your justifications are very clear -- thanks! I think you could make of them your first entry in PM, maybe attached to the "trivial valuation".
Regards,
Jussi