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. will be a number field, and an absolute value (which, by abuse of terminology, we will use interchangeable with “valuation” and “place”) extending one of the absolute values on (which are always normalized in the standard way); we will write for the output at .
Suppose is a valuation of ; we write if extends . Recall the following important formula from the theory of absolute values an extension fields:
Write ; these are the local degrees. From elementary algebraic number theory, we have This is essentially a version of the formula.
The above formalism allows us to deduce an important global relation between the absolute values for .
Theorem 1 (Product formula) If ,
The proof of this theorem starts with the case , in which case it is an immediate consequence of unique factorization. For instance, one can argue as follows. Let , a prime. Then by the standard normalization, if , and where is the standard archimedean absolute value on . The formula is thus clear for , and it follows in general by multiplicativity.
By the above reasoning, it follows that if ,
This completes the proof.
1. Counting elements in boxes
We shall be interested in studying elements with certain distributions of its valuations . In particular, let us pose the following problem.
Suppose we have constants for each place . When does there exist with for all ?
There is an analogy here to algebraic geometry. Suppose is a projective nonsingular curve. Let be a divisor, i.e. a formal finite sum . When does there exist a rational function on with for all ?
Information on this problem is provided by the Riemann-Roch theorem. In particular, if is very large, then we can always find a lot of such s. In particular, when is very large, the dimension of the space of such is approximately proportional to .
Note the strength of the analogy: the valuation rings of the function field of are precisely the local rings of the points of . Perhaps it is more than an analogy, but I’m just not informed enough to know.
In our case, we will define a -divisor to be an assignment of constants for the valuations of . We will assume that almost everywhere (i.e. for a cofinite set); this is analogous to a divisor in algebraic geometry being finite. Also, when is nonarchimedean, we assume in the value group of .
Define to be the set of with for all . We will show that is approximately proportional to
2. The main counting result
Theorem 2 There are positive constants depending on only such that
Note that we need the 1 on the right side, because 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 , then by the pigeonhole principle we could find ones that very close together. But the difference of any two points in 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 is a real archimedean absolute value of . Let . Then, by the pigeonhole principle, there are distinct points with
because if is the imbedding corresponding to , then forms a subset of of cardinality .
Then, by splitting the product into and non- valuations, and those into the archimedean and non-archimedean ones, we find:
Here the symbol means up to a constant factor depending only on . This proves one side of the equality when there is a real archimedean absolute value of .
When there are only complex ones, the proof is slightly more complicated; one finds with 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 into subsquares, where is the smallest square , and use that there must be a subsquare containing two distinct elements of .
Now, we prove the left hand inequality, bounding below . This will be the necessary part for the unit theorem. First off, note that and are unaffected by scalar multiplication by elements of (where denotes the family ), in view of the product formula. By multiplying by an element of , we may assume at the archimedean places, by the Artin-Whaples approximation theorem. Next, we choose divisible by a lot of prime numbers, so that has absolute value at all nonarchimedean primes. Replace with .
So, here is the situation. There is a satisfying for nonarchimedean and for archimedean. We need to show that there are elements in . But, has a nice interpretation in this case. There is an ideal such that the elements of are precisely the with for nonarchimedean.
If we show that the number of elements of is at least proportional to , by an absolute constant (depending only on the field ) we will be done; the normalization of the absolute values is carried out such that is, up to a constant factor determined by what happens at the archimedean places, precisely this. Here denotes the cardinality of (i.e. the ideal norm to ), where is the ring of integers in .
Let be a basis of the -module . Then if denotes the the maximum of for archimedean and , then any sum
satisfies for archimedean.
The number of such with is . Also, since there are different classes modulo , it follows that there are at least
belonging to the same class modulo . Subtracting these from each other, we get at least elements of whose norms at the archimedean places are at most . This proves that , since depends only on . And completes the proof.