The Riemann-Roch problem in number fields

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.