Geometry of Numbers

4 Distribution of ideals

We would like to count the ideals of norm \(\leq n\) in a number field. To do this, we will fix an ideal class, and let \(I\) be a representative of the inverse class. Then ideals of norm \(\leq n\) are the same as principle ideals inside \(I\) with norm \(\leq n N(I)\). To count principle ideals, we only have to count generators. Unfortunately in general, there can be a lot of units. To deal with this problem, let \(A\) be a fundamental parallelogram for a fundamental system of units \(u_i\) in the hyperplane they lie in for the log embedding, and let \(v\) be the vector that is \([F_i:\RR ]\) in each corresponding direction, i.e. the normal vector of the hyperplane \(A\) lies in. Now \(A+\RR v\) are representatives of a coset of the units in the log embedding, and so the preimage \(D\) under the log embedding up to roots of unity has a unique generator of every principle ideal. Moreover, because of how \(v\) was chosen, this region is scaling invariant.

Let \(N_n\) be the region of \(\RR ^r\times \CC ^s\) with \(N(x) \leq n\). Then the number of ideals in the ideal class is equal to the number of lattice points in \(N_{nN(I)}\cap D\) divided by the number of roots of unity. Note that \(N_{a}= a^{\frac 1 n}N_1\), so that \(D_a=N_a\cap D =a^{\frac {1}{n}} (N_1\cap D)\). We would like to prove that the number of lattice points of a lattice \(L\) in \(aD_1\) is \(\frac {A(D_1)}{A(L)}a^n+O(a^{n-1})\), but this is true for any region with reasonable boundary.

Namely, let \(\partial B\) be (n-1) Lipschitz parameterizable if it is a finite union of images of cells under Lipschitz maps.

Lemma 4.1. If \(B \subset \RR ^n\) has (\(n-1\)) Lipschitz parameterizable boundary, then \(\#\{aB \cap L\} = \frac {A(B_1)}{A(L)}a^n+O(a^{n-1})\).

Proof. First note that it suffices to assume \(L\) is the standard lattice. Let \(\ZZ ^n\) translates of \([0,1]^n\) be standard cubes. The difference between the number lattice points and the is bounded above by the number of standard cubes intersecting the boundary \(\partial B\), so we need to show this is \(O(a^{n-1})\). WLOG, \(\partial B\) can be the image of one \([0,1]^{n-1}\). To get a bound, break up each cube on the boundary into \(\floor a\) pieces in each direction, and note that if \(\lambda \) is the Lipschitz constant, then \(\frac {\sqrt {n}\lambda }{{\floor a}^{n-1}}\) is an upper bound on the maximum distance between any two points on the boundary of each subdivision, so the number of cubes hit by each subdivision is bounded above by a constant \(c\), and so the total number of cubes hit is bounded above by \(ca^{n-1}\). □

It suffices to see that \(D_1\) has a Lipschitz parameterization, and compute its volume. We will replace \(D_1\) by \(D_1^+\), the subset where each real component is required to be positive. Clearly \(A(D_1^+) = 2^{-r}A(D_1)\). If \(v_j\) are the coordinates in the log embedding of the fundmental system of units, then the fundamental domain is \(\{\sum _{k=1}^{r+s-1} t_j v_j|0 \leq t_j <1\}\). let \(v_j^k\) be the components of \(v_j\). If \(x_1,\dots x_r,z_1,\dots z_s\) are coordinates in \(\RR ^r\times \CC ^s\), then for every point \((x_1,\dots ,x_r,z_1,\dots z_s)\) in \(D_1^+\) we have \(\log x_i = \sum _{k=1}^{r+s-1}t_kv_k^i+u\) and \(2\log |z_i| = \sum _{k=1}^{r+s-1}t_kv_k^i+2u\), where \(u \leq 0\). Let \(t_{r+s} = e^u\) so that it ranges in \((0,1]\). Now can write each nonzero \(z_i\) uniquely as \(\rho _i e^{i\theta _i}\). The \(\theta _i,t_k\) give a Lipschitz parameterization of the region.

Moreover, we can use the parameterization to compute the area. Using polar coordinates, we have \(\int _{D_1^+} \prod \rho _i dx_jd\rho _id\theta _k\). We will change coordinates from the \(x_j,\rho _i\) to \(t_j\). for \(t<{r+s}\), \(\frac {dx_i}{dt_j}\) is \(x_iv_j^i\), and similarly \(\frac {d\rho _i}{dt_j}\) is \(\frac 1 2 \rho _iv_j^{r+i}\). \(\frac {dx_i}{dt_{r+s}} = x_i\), \(\frac {d\rho _i}{dt_{r+s}} = \rho _i\). The determinant of this Jacobian will be \(2^{-s}\prod x_i \prod \rho _i\) times the determinant of the coordinates of the \(v_j\) and \(v\). The determinant of \(v_j\) and \(v\) is defined to be \(n\) times the regulator of \(K\) denoted \(\reg (K)\), so after the change of coordinates we have \(\pi ^s\int _{0 \leq t_i \leq 1} \prod \rho _i^2 \prod x_i dt_1\dots dt_{r+s}\). The product in the integral is \(t_{r+s}^{n-1}\) since the \(v_i\) are units so their coordinates sum up to \(0\). We end up with \({\pi ^s}{\reg (K)}\) as our volume.

Theorem 4.2. Let \(j_K^c(n)\) be the number of ideals of norm at most \(n\) in the ideal class \(c\). Then \(j_K^c(n) = \frac {2^{r+s}\pi ^s \reg (K)}{\omega _K \sqrt {|\disc (K)|}}n + O(n^{1-\frac {1}{[K:\QQ ]}})\). If \(j_K(n)\) is the number of ideals of norm at most \(n\), then \(j_K(n) = \frac {h_K2^{r+s}\pi ^s \reg (K)}{\omega _K \sqrt {|\disc (K)|}}n + O(n^{1-\frac {1}{[K:\QQ ]}})\), where \(h_K\) is the class number, and \(\omega _K\) is the number of roots of unity.

Proof. As argued before, \(\omega _Kj_K^c(n) = \frac {A(D_1)}{A(I)}N(I)n+O(n^{1-\frac {1}{[K:\QQ ]}})\), but we have computed \(A(D_1)\), and we know \(A(I)\), so putting everything together gives the formula. □