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Let $K$ be a number field, with ring of integers $\mathcal{O}_k$. For $x\in \mathcal{O}_K$, let $f(x) = |N_{K/\mathbb{Q}}(x)|$, the (usual) absolute value of the norm of $x$ over $\mathbb{Q}$. Dirichlet's unit theorem tells us, in a precise sense, "how many" units (i.e., elements with $f(x)=1$) there are in $\mathcal{O}_K$.

My question is, are there results out there about the size (or structure) of the set of elements $\alpha\in \mathcal{O}_K$ with $f(\alpha)=n$, for some given positive integer $n$? (To make things more concrete, we could just focus on $n=2$, if it helps.)

Of course, if $f(\alpha)=n$ and $u$ is a unit, $f(u\alpha)=n$ too, so we should mod out by the action of $\mathcal{O}_K^*$ on $\mathcal{O}_K$ by multiplication, i.e., look at equivalence classes of associated elements of $\mathcal{O}_K$. If we let $S =$ this set of equivalence classes, $f$ is well-defined on $S$. For a given $n\in\mathbb{N}$, is $f^{-1}(n)$ finitely generated or (dare I hope it) finite?

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3 Answers 3

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Okay here is part of an answer to your question:

Let $x \in O_K$. Then it can be shown that the norm of $x$ is the norm of the ideal $(x)$ (I am presuming you know what the norm of an ideal is. If not, take a look at Pierre Samuel's book "Algebraic Theory of Numbers"). But, since $O_K$ is a Dedekind domain, it can be shown (using an argument of unique factorization of ideals into products of primes) that there are only finitely many integral ideals in $O_K$ of a given norm. Thus, up to units, there will be only finitely many elements in $O_K$ of a given norm.

Here is the reason why there are only finitely many ideals of norm $n \in \mathbb{N}$:

Let $I \subset O_K$ be an ideal of norm $n$. Then by definition of the norm of an ideal, $|O_K/I| = n$. Thus, $n \in I$. This means that $(n)O_K \subset I$. Factorize $(n)O_K$ into a product of prime ideals. Then one can see that there are only finitely many choices for $I$.

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I should mention that this is a standard proof, which you can read in most ANT textbooks. I feel bad, because I have left out details in the proof. If any part of my argument is unclear to you, then read up some relevant facts about factorization of ideals in Dedekind domains. –  Rankeya Dec 18 '11 at 4:12
    
Don't feel bad! This answers my basic question of whether it's finite. –  jdc Dec 18 '11 at 4:25
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You don't need to know anything about prime ideals to prove finiteness. Just note that $n\mathcal{O}_K\subseteq I\subseteq \mathcal{O}_K$, so it's enough to prove finiteness of $\mathcal{O}_K/n\mathcal{O}_K$. But $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $d=$ degree of $K$, so that quotient is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^d$ as an abelian group, and the ideal $I$ corresponds to a subgroup of that. The multiplicative structure of $I$ only enters in a very weak way: all you need to know is that $I$ is a subgroup of $\mathcal{O}_K$ of full $\mathbb{Z}$-rank. –  Alex B. Dec 18 '11 at 4:38
    
Well, that is true, but if you are studying ANT, it is good to know about factorization of ideals in Dedekind domains. –  Rankeya Dec 18 '11 at 4:40

Counting elements of a given norm modulo units is the same as counting principal ideals of a given norm. It's a quite difficult problem. For example, how many elements of norm $N$ are there in $\mathbf{Z}[i]$? In other words, in how many ways can we write $N$ as the sum of two squares? A theorem of Jacobi says that this is equal to four times the excess of the number of divisors of $N$ which are $\equiv 1 \mod 4$ over the number of divisors of $N$ which are $\equiv 3 \mod 4$, generalizing the theorem of Euler-Fermat about representations of primes as sums of two squares. Thus we see that for even the simplest number field $\mathbf{Q}(i)$, the situation is quite complicated (but in this case, Jacobi's theorem provides a simple formula).

The number $a_N$ of ideals of norm $N$ in $\mathcal{O}_K$ is related to the residue of the Dedekind zeta function $\zeta(s)$ at $s=1$, which is in turn related to all important invariants of $K$.

We can write

$$\zeta_K(s)=\sum_{n \geq 1}\frac{a_n}{n^s},$$

where the sum converges absolutely for $\Re s>1$.

According to the Wiener–Ikehara Tauberian theorem, since $\zeta_K$ has a simple pole at $s=1$,

$$\sum_{n\leq N}a_N \sim C_K N$$

where $C_K$ is the residue of $\zeta_K$ at $s=1$, given explicitly by the class number formula.

This gives an upper bound for the number of principal ideals of norm at most $N$, and an asymptotic expression in case $\mathcal{O}_K$ is a P.I.D.

Addendum: As pointed out by Matt, ideals are uniformly distributed among the $h$ ideal classes of $\mathcal{O}_K$. This is the heart of the class number formula.

By definition of the ideal class group of $\mathcal{O}_K$, the set of principal ideals of $\mathcal{O}_K$ is the identity element of the ideal class group. By the uniformity of the distribution, if we denote $L$ the average over $n$ of the number of principal ideals of norm $n$ as $n \to \infty$, then

$$\frac{1}{h}\sum_{n\leq N}a_n \sim \frac{1}{h}C_KN \sim LN$$

hence $$L=\frac{1}{h}C_K = \frac{2^{r_1}(2\pi)^{r_2}\text{Reg}_K}{\omega_K \sqrt{|D_K|}}$$

where $r_1$ and $r_2$ are respectively the number of real and complex infinite primes of $K$, $\text{Reg}$ is the regulator (the covolume of the unit group $\mathcal{O}_K^*$ in $K^*$), $D_K$ is the discriminant (more or less the square of the covolume of $\mathcal{O}_K$ in $K$), and $\omega_K$ is the number of roots of unity in $K^*$.

Addendum #2: I just want to point out what I think is the neatest proof of of Jacobi's theorem, using $L$-functions.

We let $K=\mathbb{Q}(i)$. By definition of the $\zeta_K(s)$, and using the fact that there are two primes of norm $p$ above $p \equiv 1 \mod 4$, one prime of norm $p^2$ above $p \equiv 3 \mod 4$, and one prime of norm $2$ above $p=2$,

$$\zeta_K(s)= (1-2^{-s})^{-1}\prod_{p \equiv 3 \mod 4} (1-p^{-2s})^{-1} \prod_{p \equiv 1 \mod 4} (1-p^{-s})^{-2}$$

which, be rearranging, can be written as $\zeta(s)L(s, \chi)$, where $\chi$ is the Dirichlet character $$p \mapsto \begin{cases}\left(\frac{-1}{p}\right) = (-1)^{(p-1)/2} && p \text{ odd}; \\ 0 && p=2\end{cases}.$$

Hence, from this identity, the coefficient of $N^{-s}$ in $\zeta_K$ is

$$a_N = \sum_{d \mid N}\chi(d),$$

which is precisely what Jacobi's theorem says, since for a general odd integer $n$ we have $\chi(n)=(-1)^{(n-1)/2} = \pm 1$ according as to $n\equiv 1$ or $n \equiv 3$ mod $4$.

$$$$

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Dear Bruno, By considering the $L$-functions $L(\chi,s)$ as $\chi$ ranges over the various characters of the ideal class group, can't we deduce that for each ideal class $c$, the sum $\sum_{\mathfrak a \in c} N(\mathfrak a) \simeq \dfrac{1}{h} C_K N$? Regards, –  Matt E Dec 18 '11 at 5:07
    
Dear @MattE, you are right! I will add it immediately. Thank you! –  Bruno Joyal Dec 18 '11 at 5:19
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Dear Bruno, You're welcome. Regards, –  Matt E Dec 18 '11 at 7:03
    
Rankeya answered my basic question of finiteness, but thank you for the additional information! Lovely stuff. –  jdc Dec 19 '11 at 17:32
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The post has been un-wikified. –  Zev Chonoles Dec 28 '11 at 15:01

This is an addendum to Addendum 2 in Bruno's great answer.

Let $d$ be a square free integer, let $K$ be the quadratic field $\mathbb Q(\sqrt d)$, let $\mathbb Z_K$ be its ring of integers, and let $D$ be the integer which equal to $d$ if $d\equiv1\bmod4$, and to $4d$ otherwise.

Following

Borevich-Shafarevich, Number Theory, p. 237,

Borevitch-Chafarevitch, Théorie des Nombres, p. 263,

define $\chi:\mathbb Z\to\mathbb Z$ as follows. If $x$ is not prime to $D$, put $\chi(x)=0$. If $x$ is prime to $D$, let

$$d\equiv1\bmod4\implies\chi(x)=\left(\frac{x}{|d|}\right),$$

$$d\equiv3\bmod4\implies\chi(x)=(-1)^{(x-1)/2}\left(\frac{x}{|d|}\right),$$

$$d=2d'\implies\chi(x)=(-1)^{[(x^2-1)/8]+[(x-1)/2]+[(d'-1)/2]}\left(\frac{x}{|d'|}\right),$$ where $(\frac{a}b)$ is the Jacobi symbol.

Then $\chi$ induces a multiplicative map from $\mathbb Z/D\mathbb Z$ to $\mathbb Z$. Moreover, if $p$ is prime, then

$\bullet$ $\chi(p)=1$ if $p\mathbb Z_K$ is the product of two prime ideals of norm $p$,

$\bullet$ $\chi(p)=-1$ if $p\mathbb Z_K$ is a prime ideal of norm $p^2$,

$\bullet$ $\chi(p)=0$ if $p\mathbb Z_K$ is the square of a prime ideal of norm $p$.

This implies $$ \zeta_K(s)=\zeta(s)\ L(s,\chi), $$ that is $$ \text{(Dedekind)}=\text{(Riemann)}\cdot\text{(Dirichlet)}, $$ and the number of ideals of $\mathbb Z_K$ of norm $n$ is $$ \sum_{m|n}\ \chi(m). $$ Now assume $d < 0$.

$(*)$ The number $u$ of units of $\mathbb Z_K$ is $4$ if $d=-1$, it is $6$ if $d=-3$, and it is $2$ otherwise.

See for instance this entry of PlanetMath.

Suppose in addition that $\mathbb Z_K$ is principal, i.e. that $d$ is one of the numbers

$$-1, -2, -3, -7, -11, -19, -43, -67, -163.$$

See for instance this sub-entry of Wikipedia.

The number of elements of norm $n$ is then $$ u\ \sum_{m|n}\ \chi(m), $$ where $u$ is the number of units of $\mathbb Z_K$, see $(*)$.

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