Suppose is a finite extension of fields, and that and are valuations on and , respectively.

*Uniqueness.* View as a
finite-dimensional vector space over . Then
is a norm in
the sense defined earlier (Definition 16.1.1). Hence any two
extensions
and
of
are equivalent
as norms, so induce the same topology on . But as we have
seen (Proposition 14.1.4), two valuations which induce the same topology are
equivalent valuations, i.e.,
, for some
positive real . Finally since
for all .

*Existence.* We do not give a proof of
existence in the general case. Instead we give a proof, which was
suggested by Dr. Geyer at the conference out of which
[Cas67] arose. It is valid when is locally
compact, which is the only case we will use later.

We see at once that the function defined in (17.1.1) satisfies the condition (i) that with equality only for , and (ii) for all . The difficult part of the proof is to show that there is a constant such that

Choose a basis for over . Let be the max norm on , so for with we have

With respect to the -topology, has the product topology as a product of copies of . The function is a composition of continuous functions on with respect to this topology (e.g., is the determinant, hence polynomial), hence defines nonzero continuous function on the compact set

for all $a&isin#in;S$

For any nonzero there exists such that
; to see this take to be a
in the expression
with
for any . Hence
, so and

(say) |

as required.

for all $a&isin#in;M$

But now

as required.

I don't believe this proof, which I copied from Cassels's article. My problem with it is that the proof of Theorem 17.1.2 does not give that , i.e., that the triangle inequality holds for . By changing the basis for one can make any nonzero vector have , so if we choose such that is very large, then the in the proof will also be very large. One way to fix the corollary is to only claim that there are positive constants such that

When is no longer complete under the position is more complicated:

It remains to show that the are distinct and that they are the only extensions of to .

Suppose is any valuation of that extends . Then extends by continuity to a real-valued function on , which we also denote by . (We are again using that is dense in .) By continuity we have for all ,

We consider the restriction of to one of the . If for some , then for every in so . Hence either is identically 0 on or it induces a valuation on .

Further, cannot induce a valuation on two of the . For

It remains only to show that (17.1.2) is a topological homomorphism. For

William Stein 2012-09-24