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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
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| ``Re: 6.''
by lance on 2006-12-15 17:54:37 |
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| 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. |
| | [ reply | up | top ] |
- Re: 6. by pahio on 2006-12-18 13:03:29
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