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The Artin-Whaples approximation theorem October 6, 2009

Posted by Akhil Mathew in algebra, algebraic number theory, number theory.
Tags: , ,

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 {\left|\cdot\right|_1, \dots, \left|\cdot\right|_n} on the field {K}; this means that they induce different topologies, so are not powers of each other

Theorem 1 (Artin-Whaples)

Hypotheses as above, given {a_1, \dots, a_n \in K} and {\epsilon>0}, there exists {a \in K} with


\displaystyle \left|a - a_i\right|_i < \epsilon, \quad 1 \leq i \leq n.



I claim that first of all, it is sufficient to take the case where {a_1 = 1, a_i = 0} for {i>1}. Indeed, if this case is proven, then (by symmetry) choose for each {i}, {b_i} approximating {1} at {\left|\cdot\right|_i} and {0} elsewhere, and take

\displaystyle a = \sum a_i b_i.

Before proving the approximation theorem, we prove:

Lemma 2 Hypotheses as above, there exists {a \in K} with {\left|a\right|_1 > 1} and {\left|a\right|_i <1} if {i>1} 

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

By induction on {n}, assume there is {a' \in K} with {\left|a'\right|_1 > 1} and {\left|a'\right|_1 < 1} if {1<i<n}

Case 1

If {\left|a'\right|_n < 1}, then we’re already done with {a=a'}

Case 2

If {\left|a'\right|_n > 1}, consider

\displaystyle \frac{a'^N}{1 + a'^N}

which when {N} is large is close to 1 at {\left|\cdot\right|_1, \left|\cdot\right|_n} and close to zero elsewhere. By the case {n=2}, there is {a'' \in K} with {\left|a''\right|_1 <1} and {\left|a''\right|_n > 1}. Then take {a = a'' \frac{a'^N}{1 + a'^N} }.  

Case 3

If {\left|a'\right|_n = 1}, then choose {a''} as in Case 2 and let {a = a'' a'^N}, where {N} is large enough to bring the absolute values at {\left|\cdot\right|_i}, {1 Mi<n}, down below 1.

Now finally we have to establish the theorem in the case we described, where {a_1 = 1, a_i = 1} if {i>1}. For this, simply take a high power of {a}


Ok, now why do we care? 

Theorem 3

Let {K} be a field with a discrete valuation {v} (yes, this is a change of notation), and let {L} be a finite extension with {w} the valuations on {L} prolonging {v}. Then


\displaystyle \prod_w L_w \simeq L \otimes_K K_v.



Here {K_v} (resp. {L_w}) denotes the completion at {v} (resp. {w}).

First of all, there is a {K}-linear inclusion map {L \rightarrow \prod_w L_w}, which induces a {K_v}-linear map

\displaystyle L \otimes_K K_v \rightarrow \prod_w L_w .

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

The image is dense by the Artin-Whaples theorem. Also, the dimensions of the images map. If the left is {n=[L:K]}, the right is

\displaystyle \sum_{w |v} [L_w: K_v] = \sum e_w f_w,

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

But we know that {n = \sum e_w f_w}. 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 {V \rightarrow V'} 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 {V} 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 {V'} 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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