This post won’t be as cool as the title sounds. But I will prove something neat, and it will lead to neater things as time goes on (assuming I keep posting on this topic).

We will now return to algebraic number theory, following Lang’s textbook, and study the distribution of points in parallelotopes.

The setup is as follows. {K} will be a number field, and {v} an absolute value (which, by abuse of terminology, we will use interchangeable with “valuation” and “place”) extending one of the absolute values on {\mathbb{Q}} (which are always normalized in the standard way); we will write {|x|_v} for the output at {x \in K}.

Suppose {v_0} is a valuation of {\mathbb{Q}}; we write {v | v_0} if {v} extends {v_0}. Recall the following important formula from the theory of absolute values an extension fields:

\displaystyle |N^K_{\mathbb{Q}}(x)|_{v_0} = \prod_{v | v_0} |x|_v^{ [K_v: \mathbb{Q}_{v_0} ] }.

Write {N_v := [K_v: \mathbb{Q}_{v_0} ] }; these are the local degrees. From elementary algebraic number theory, we have {\sum_{v | v_0} N_v = N := [K:\mathbb{Q}].} This is essentially a version of the {\sum ef = N} formula.

The above formalism allows us to deduce an important global relation between the absolute values {|x|_v} for {x \in K}.

Theorem 1 (Product formula) If {x \neq 0},\displaystyle \prod_v |x|_v^{N_v} = 1.

 

The proof of this theorem starts with the case {K=\mathbb{Q}}, in which case it is an immediate consequence of unique factorization. For instance, one can argue as follows. Let {x = p}, a prime. Then {|p|_p = 1/p} by the standard normalization, {|p|_l = 1} if {l \neq p}, and {|p|_{\infty} = 1} where {|\cdot|_\infty} is the standard archimedean absolute value on {\mathbb{Q}}. The formula is thus clear for {x = p \in \mathbb{Q}}, and it follows in general by multiplicativity.

By the above reasoning, it follows that if {x \in K^*},

\displaystyle \prod_v |x|_v^{N_v} = \prod_{v_0} |N^K_{\mathbb{Q}}x|_{v_0} = 1.

This completes the proof.

1. Counting elements in boxes

We shall be interested in studying elements {x \in K} with certain distributions of its valuations {|x|_v}. In particular, let us pose the following problem.

Problem

Suppose we have constants {C_v} for each place {v}. When does there exist {x \in K^*} with {|x|_v \leq C_v} for all {v}?

There is an analogy here to algebraic geometry. Suppose {C} is a projective nonsingular curve. Let {D} be a divisor, i.e. a formal finite sum {\sum_{P \in C} n_P P}. When does there exist a rational function {f} on {C} with {ord_{P}(f) \geq -n_P} for all {P}?

Information on this problem is provided by the Riemann-Roch theorem. In particular, if {|D| := \sum n_P} is very large, then we can always find a lot of such {f}s. In particular, when {|D|} is very large, the dimension of the space {L(D)} of such {f} is approximately proportional to {|D|}.

Note the strength of the analogy: the valuation rings of the function field of {C} are precisely the local rings of the points of {C}. Perhaps it is more than an analogy, but I’m just not informed enough to know.

In our case, we will define a {K}-divisor {\mathfrak{c}} to be an assignment of constants {C_v} for the valuations {v} of {K}. We will assume that {C_v = 1} almost everywhere (i.e. for a cofinite set); this is analogous to a divisor in algebraic geometry being finite. Also, when {v} is nonarchimedean, we assume {C_v} in the value group of {v}.

Define {L(\mathfrak{c})} to be the set of {x \in K} with {|x|_v \leq C_v} for all {v}. We will show that {\mathrm{card} L(\mathfrak{c})} is approximately proportional to

\displaystyle \left \lVert{\mathfrak{c}} \right \rVert := \prod C_v^{N_v}.

2. The main counting result

Theorem 2 There are positive constants {a,b} depending on {K} only such that\displaystyle a \left \lVert{c}\right \rVert \leq \mathrm{card} L(\mathfrak{c}) \leq b(1 +\left \lVert{c} \right \rVert).

 

Note that we need the 1 on the right side, because {0 \in L(\mathfrak{c})} always!

The proof of this theorem will not use any of the fancy machinery as in the Riemann-Roch theorem (in particular, no Serre duality). It only gives bounded proportionality; one can prove a stronger asymptotic result, but we shall not do so here.

We first do the right inequality. The idea is simple. If there were a lot of points in {L(\mathfrak{c})}, then by the pigeonhole principle we could find ones that very close together. But the difference of any two points in {L(\mathfrak{c})} has bounded absolute values everywhere, so if the difference is very small at even one valuation, then the product formula will be violated.

In detail, it is as follows. Suppose {w} is a real archimedean absolute value of {K}. Let {n = \mathrm{card} L(\mathfrak{c})}. Then, by the pigeonhole principle, there are distinct points {x,y \in L(\mathfrak{c})} with

\displaystyle |x-y|_w \leq 2 C_{w}/n ;

because if {\sigma: K \rightarrow \mathbb{R}} is the imbedding corresponding to {w}, then {\sigma L(\mathfrak{c})} forms a subset of {[-C_w, C_w]} of cardinality {n}.

Then, by splitting the product into {w} and non-{w} valuations, and those into the archimedean and non-archimedean ones, we find:

\displaystyle 1 = \prod |x-y|_v^{N_v} = \frac{ 2 C_w}{n} \prod_{v \neq w }^{ \text{arch}} 2 C_v \prod_{v \neq w}^{ \text{nonarch}} C_v \lesssim \left \lVert \mathfrak{c}\right \rVert /n.

Here the symbol {\lesssim} means {\leq} up to a constant factor depending only on {K}. This proves one side of the equality when there is a real archimedean absolute value of {K}.

When there are only complex ones, the proof is slightly more complicated; one finds {x,y} with {|x-y|_w \leq 2 \sqrt{2} C_w/\sqrt{n}} by using a bit of 2-dimensional geometry (but the idea is the same). For instance, we could cut the square centered at zero fo side length {2C_w} into {m^2} subsquares, where {m^2} is the smallest square {< n}, and use that there must be a subsquare containing two distinct elements of {L(\mathfrak{c})}.

Now, we prove the left hand inequality, bounding below {\mathrm{card} L(\mathfrak{c})}. This will be the necessary part for the unit theorem. First off, note that {\mathrm{card} L(\mathfrak{c})} and {\left \lVert \mathfrak{c}\right \rVert} are unaffected by scalar multiplication by elements of {K} (where {x \mathfrak{c}} denotes the family {|x|_v C_v}), in view of the product formula. By multiplying {\mathfrak{c}} by an element of {K}, we may assume {1 \leq C_v \leq 2} at the archimedean places, by the Artin-Whaples approximation theorem. Next, we choose {d \in \mathbb{Z}} divisible by a lot of prime numbers, so that {d \mathfrak{c}} has absolute value {\leq 1} at all nonarchimedean primes. Replace {\mathfrak{c}} with {d \mathfrak{c}}.

So, here is the situation. There is a {\mathfrak{c}} satisfying {C_v \leq 1} for {v} nonarchimedean and {d \leq C_v \leq 2d} for {v} archimedean. We need to show that there are {\gtrsim \left \lVert \mathfrak{c}\right \rVert} elements in {L(\mathfrak{c})}. But, {L(\mathfrak{c})} has a nice interpretation in this case. There is an ideal {\mathfrak{a}= \prod \mathfrak{p}^{\mathrm{ord}_{\mathfrak{p}} C_{\mathfrak{p}}}} such that the elements of {\mathfrak{a}} are precisely the {x \in K} with {|x|_v \leq C_v} for {v} nonarchimedean.

If we show that the number of elements of {\mathfrak{a}} is at least proportional to {d^N/ \mathbf{N} \mathfrak{a}}, by an absolute constant (depending only on the field {K}) we will be done; the normalization of the absolute values is carried out such that {\left \lVert \mathfrak{c} \right \rVert} is, up to a constant factor determined by what happens at the archimedean places, precisely this. Here {\mathbf{N}\mathfrak{a}} denotes the cardinality of {\mathcal{O}/\mathfrak{a}} (i.e. the ideal norm to {\mathbb{Q}}), where {\mathcal{O}} is the ring of integers in {K}.

Let {x_1, \dots, x_N} be a basis of the {\mathbb{Z}}-module {\mathcal{O}}. Then if {W} denotes the the maximum of {N|x_i|_w} for {w} archimedean and {1 \leq i \leq N}, then any sum

\displaystyle x = \sum a_i x_i , \forall a_i \in \mathbb{Z}, |a_i| \leq d/W

satisfies {|x|_w \leq d} for {w} archimedean.

The number of such {x} with {0 \leq a_i \leq d/W} is {(d/W)^N}. Also, since there are {\mathbf{N}\mathfrak{a}} different classes modulo {\mathfrak{a}}, it follows that there are at least

\displaystyle \frac{d^N}{W^N \mathbf{N} \mathfrak{a}}

belonging to the same class modulo {\mathfrak{a}}. Subtracting these from each other, we get at least {\frac{d^N}{W^N \mathbf{N} \mathfrak{a}} } elements of {\mathfrak{a}} whose norms at the archimedean places are at most {d}. This proves that {\mathrm{card} L(\mathfrak{c}) \gtrsim \left \lVert{\mathfrak{c}}\right \rVert}, since {W^N} depends only on {K}. And completes the proof.

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