The Artin-Whaples approximation theorem is a nice extension of the Chinese remainder theorem to absolute values, to which it reduces when the absolute values are discrete.

So fix pairwise nonequivalent absolute values on the field ; this means that they induce different topologies, so are not powers of each other.

Theorem 1 (Artin-Whaples)Hypotheses as above, given and , there exists with

I claim that first of all, it is sufficient to take the case where for . Indeed, if this case is proven, then (by symmetry) choose for each , approximating at and elsewhere, and take

Before proving the approximation theorem, we prove:

Lemma 2Hypotheses as above, there exists with and if .

The case of two absolute values follows from nonequivalence; see the previous post.

By induction on , assume there is with and if .

**Case 1 **

If , then we’re already done with .

**Case 2 **

If , consider

which when is large is close to 1 at and close to zero elsewhere. By the case , there is with and . Then take .

**Case 3 **

If , then choose as in Case 2 and let , where is large enough to bring the absolute values at , , down below 1.

Now finally we have to establish the theorem in the case we described, where if . For this, simply take a high power of .

**Applications **

Ok, now why do we care?

Theorem 3Let be a field with a discrete valuation (yes, this is a change of notation), and let be a finite extension with the valuations on prolonging . Then

Here (resp. ) denotes the completion at (resp. ).

First of all, there is a -linear inclusion map , which induces a -linear map

Note that both sides are finite-dimensional vector spaces over the complete field .

The image is dense by the Artin-Whaples theorem. Also, the dimensions of the images map. If the left is , the right is

because of the following. is the product of and because of the remarks closing yesterday’s post. Note also that and don’t change when one passes to the completion—this is because depends on the value group, which completion doesn’t change (by nonarchimedeanness), and by a density argument. (In general, even for rings, completion w.r.t. a maximal ideal preserves the residue field.)

But we know that . Consequently the two spaces are of equal dimension and have a dense map between the two; this dense map is thus bijective. (To be precise, if is a map of finite-dimensional vector spaces over a complete field if if V is a finite-dimensional vector space over a complete field, the density of the image of implies that the map is surjective. density of a subspace of V implies that it fills all of V. To see this, use the fact that all norms on are equivalent, and pick a convenient one.)

The original Artin-Whaples paper is available freely online. There’s more material in there though.

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