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, givenand
, 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 2 Hypotheses 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 3
Letbe 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.
Leave a Reply