Extensions of Valuations

In this section we continue to tacitly assume that all valuations are nontrivial. We do not assume all our valuations satisfy the triangle

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

Definition 17.1.1 (Extends)   We say that extends if for all .

Theorem 17.1.2   Suppose that is a field that is complete with respect to and that  is a finite extension of  of degree . Then there is precisely one extension of to , namely

 (17.1)

where the th root is the non-negative real th root of the nonnegative real number .

Proof. We may assume that is normalized so as to satisfy the triangle inequality. Otherwise, normalize so that it does, prove the theorem for the normalized valuation , then raise both sides of (17.1.1) to the power . In the uniqueness proof, by the same argument we may assume that also satisfies the triangle inequality.

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

Note that we do not know (and will not show) that as defined by (17.1.1) is a norm as in Definition 16.1.1, since showing that is a norm would entail showing that it satisfies the triangle inequality, which is not obvious.

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

(Note: in Cassels's original article he let be any norm, but we don't because the rest of the proof does not work, since we can't use homogeneity as he claims to do. This is because it need not be possible to find, for any nonzero some element such that . This would fail, e.g., if for any .) The rest of the argument is very similar to our proof from Lemma 16.1.3 of uniqueness of norms on vector spaces over complete fields.

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

By compactness, there are real numbers such that

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

Then by homogeneity

Suppose now that . Then , so

 (say)

as required.

Example 17.1.3   Consider the extension of equipped with the archimedean valuation. The unique extension is the ordinary absolute value on :

Example 17.1.4   Consider the extension of equipped with the -adic absolute value. Since is irreducible over we can do some computations by working in the subfield of .

Remark 17.1.5   Geyer's existence proof gives (17.1.1). But it is perhaps worth noting that in any case (17.1.1) is a consequence of unique existence, as follows. Suppose is as above. Suppose is a finite Galois extension of that contains . Then by assumption there is a unique extension of to , which we shall also denote by . If , then

is also an extension of to , so , i.e.,

for all $a&isin#in;M$

But now

for , where extend the embeddings of into . Hence

as required.

Corollary 17.1.6   Let be a basis for over . Then there are positive constants and such that

for any not all 0.

Proof. For and are two norms on considered as a vector space over .

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

Then choose such that and satisfies the triangle inequality, and prove the modified corollary using the proof suggested by Cassels.

Corollary 17.1.7   A finite extension of a completely valued field is complete with respect to the extended valuation.

Proof. By the proceeding corollary it has the topology of a finite-dimensional vector space over . (The problem with the proof of the previous corollary is not an issue, because we can replace the extended valuation by an inequivalent one that satisfies the triangle inequality and induces the same topology.)

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

Theorem 17.1.8   Let be a separable extension of of finite degree . Then there are at most extensions of a valuation on to , say , for . Let be the completion of with respect to , and for each  let be the completion of with respect to . Then

 (17.2)

algebraically and topologically, where the right hand side is given the product topology.

Proof. We already know (Lemma 16.2.1) that is of the shape (17.1.2), where the are finite extensions of . Hence there is a unique extension of to the , and by Corollary 17.1.7 the are complete with respect to the extended valuation. Further, the ring homomorphisms

are injections. Hence we get an extension of to by putting

Further, is dense in with respect to because is dense in (since is dense in ). Hence is exactly the completion of .

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 ,

and if is the constant in axiom (iii) for and , then

(In Cassels, he inexplicable assume that at this point in the proof.)

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

so for any , ,

Hence induces a valuation in precisely one of the , and it extends the given valuation of . Hence for precisely one .

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

put

Then is a norm on the right hand side of (17.1.2), considered as a vector space over and it induces the product topology. On the other hand, any two norms are equivalent, since is complete, so induces the tensor product topology on the left hand side of (17.1.2).

Corollary 17.1.9   Suppose , and let be the minimal polynomial of . Suppose that

in , where the are irreducible. Then , where is a root of .

William Stein 2012-09-24