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

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, 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 ${a}$

Applications

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.