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Hometrivial valuation

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# trivial valuation

The trivial valuation of a field $K$ is the Krull valuation $|\cdot|$ of $K$ such that $|0|=0$ and $|x|=1$ for other elements $x$ of $K$.

# 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 $a$ be the primitive element of the multiplicative group of the field, which is cyclic. If $|\cdot|$ is any valuation of the field, then one must have $|a|=1$ since otherwise $|1|\neq 1$. Consequently, $|x|=|a^{m}|=|a|^{m}=1^{m}=1$ for all non-zero elements $x$.)

6. Every algebraic extension of finite fields has only the trivial valuation, but every field of characteristic 0 has non-trivial valuations.

## Mathematics Subject Classification

12J20*no label found*11R99

*no label found*

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## 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