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trivial valuation

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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?

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

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.

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

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