Finiteness of the Class Group via Geometry of Numbers

We have seen examples in which $ \O _K$ is not a unique factorization domain. If $ \O _K$ is a principal ideal domain, then it is a unique factorization domain, so it is of interest to understand how badly $ \O _K$ fails to be a principal ideal domain. The class group of $ \O _K$ measures this failure. As one sees in a course on Class Field Theory, the class group and its generalizations also yield deep insight into the possible abelian Galois extensions of $ K$.

Definition 10.4.1 (Class Group)   Let $ \O _K$ be the ring of integers of a number field $ K$. The $ C_K$ of $ K$ is the group of nonzero fractional ideals modulo the sugroup of principal fractional ideals $ (a)$, for $ a\in K$.

Note that if we let $ \Div (K)$ denote the group of nonzero fractional ideals, then there is an exact sequence

$\displaystyle 0 \to \O _K^* \to K^* \to \Div (K) \to C_K \to 0.

A basic theorem in algebraic number theory is that the class group $ C_K$ is finite, which follows from the first part of the following theorem and the fact that there are only finitely many ideals of norm less than a given integer.

Theorem 10.4.2 (Finiteness of the Class Group)   Let $ K$ be a number field. There is a constant $ C_{r,s}$ that depends only on the number $ r$, $ s$ of real and pairs of complex conjugate embeddings of $ K$ such that every ideal class of $ \O _K$ contains an integral ideal of norm at most $ C_{r,s}\sqrt{\vert d_K\vert}$, where $ d_K=\Disc (\O _K)$. Thus by Lemma 10.3.5 the class group $ C_K$ of $ K$ is finite. One can choose $ C_{r,s}$ such that every ideal class in $ C_K$ contains an integral ideal of norm at most

$\displaystyle \sqrt{\vert d_K\vert}\cdot \left(\frac{4}{\pi}\right)^s\frac{n!}{n^n}.

The explicit bound in the theorem is called the Minkowski bound, and I think it is the best known unconditional general bound (though there are better bounds in certain special cases).

Before proving Theorem 10.4.2, we prove a few lemmas. The strategy of the proof will be to start with any nonzero ideal $ I$, and prove that there is some nonzero $ a\in K$, with very small norm, such that $ aI$ is an integral ideal. Then $ \Norm (aI)=\Norm _{K/\mathbf{Q}}(a)\Norm (I)$ will be small, since $ \Norm _{K/\mathbf{Q}}(a)$ is small. The trick is to determine precisely how small an $ a$ we can choose subject to the condition that $ aI$ be an integral ideal, i.e., that $ a\in I^{-1}$.

Let $ S$ be a subset of $ V=\mathbf{R}^n$. Then $ S$ is if whenever $ x,y\in S$ then the line connecting $ x$ and $ y$ lies entirely in $ S$. We say that $ S$ is if whenever $ x\in S$ then $ -x\in S$ also. If $ L$ is a lattice in $ V$, then the of $ V/L$ is the volume of the compact real manifold $ V/L$, which is the same thing as the absolute value of the determinant of any matrix whose rows form a basis for $ L$.

Lemma 10.4.3 (Blichfeld)   Let $ L$ be a lattice in $ V=\mathbf{R}^n$, and let $ S$ be a bounded closed convex subset of $ V$ that is symmetric about the origin. Assume that $ \Vol (S)\geq 2^n \Vol (V/L)$. Then $ S$ contains a nonzero element of $ L$.

Proof. First assume that $ \Vol (S)>2^n\cdot \Vol (V/L)$. If the map $ \pi: \frac{1}{2}S \to V/L$ is injective, then

$\displaystyle \frac{1}{2^n}\Vol (S) = \Vol \left(\frac{1}{2} S\right)\leq \Vol (V/L),$

a contradiction. Thus $ \pi$ is not injective, so there exist $ P_1\neq P_2\in \frac{1}{2}S$ such that $ P_1-P_2\in L$. By symmetry $ -P_2\in \frac{1}{2}S$. By convexity, the average $ \frac{1}{2}(P_1-P_2)$ of $ P_1$ and $ -P_2$ is also in $ \frac{1}{2}S$. Thus $ 0\neq P_1-P_2 \in S\cap L$, as claimed.

Next assume that $ \Vol (S) = 2^n\cdot \Vol (V/L)$. Then for all $ \varepsilon >0$ there is $ 0\neq Q_\varepsilon \in L\cap (1+\varepsilon ) S$, since $ \Vol ((1+\varepsilon )S)>\Vol (S)=2^n\cdot \Vol (V/L)$. If $ \varepsilon <1$ then the $ Q_\varepsilon $ are all in $ L\cap {} 2 S$, which is finite since $ 2S$ is bounded and $ L$ is discrete. Hence there exists $ Q=Q_\varepsilon \in L\cap {} (1+\varepsilon ) S$ for arbitrarily small $ \varepsilon $. Since $ S$ is closed, $ Q\in L\cap S$. $ \qedsymbol$

Lemma 10.4.4   If $ L_1$ and $ L_2$ are lattices in $ V$, then

$\displaystyle \Vol (V/L_2) = \Vol (V/L_1) \cdot [L_1:L_2].

Proof. Let $ A$ be an automorphism of $ V$ such that $ A(L_1)=L_2$. Then $ A$ defines an isomorphism of real manifolds $ V/L_1\to V/L_2$ that changes volume by a factor of $ \vert{\mathrm{Det}}(A)\vert=[L_1:L_2]$. The claimed formula then follows. $ \qedsymbol$

Fix a number field $ K$ with ring of integers $ \O _K$. Let $ \sigma:K \to V=\mathbf{R}^n$ be the embedding

$\displaystyle \sigma(x) = \big($ $\displaystyle \sigma_1(x), \sigma_2(x),\ldots, \sigma_r(x),$    
     Re$\displaystyle (\sigma_{r+1}(x)), \ldots,$   Re$\displaystyle (\sigma_{r+s}(x)),$   Im$\displaystyle (\sigma_{r+1}(x)), \ldots,$   Im$\displaystyle (\sigma_{r+s}(x))\big),$    

where $ \sigma_1,\ldots, \sigma_r$ are the real embeddings of $ K$ and $ \sigma_{r+1},\ldots, \sigma_{r+s}$ are half the complex embeddings of $ K$, with one representative of each pair of complex conjugate embeddings. Note that this $ \sigma$ is not exactly the same as the one at the beginning of Section 10.2.

Lemma 10.4.5  

$\displaystyle \Vol (V/\sigma(\O _K)) = 2^{-s} \sqrt{\vert d_K\vert}.

Proof. Let $ L=\sigma(\O _K)$. From a basis $ w_1,\ldots,w_n$ for $ \O _K$ we obtain a matrix $ A$ whose $ i$th row is

$\displaystyle (\sigma_1(w_i), \cdots, \sigma_r(w_i),$   Re$\displaystyle (\sigma_{r+1}(w_i)),\ldots,$   Re$\displaystyle (\sigma_{r+s}(w_1)),$   Im$\displaystyle (\sigma_{r+1}(w_i)),\ldots,$   Im$\displaystyle (\sigma_{r+s}(w_1)))

and whose determinant has absolute value equal to the volume of $ V/L$. By doing the following three column operations, we obtain a matrix whose rows are exactly the images of the $ w_i$ under all embeddings of $ K$ into $ \mathbf{C}$, which is the matrix that came up when we defined $ d_K$.
  1. Add $ i=\sqrt{-1}$ times each column with entries Im$ (\sigma_{r+j}(w_i))$ to the column with entries Re$ (\sigma_{r+j}(w_i))$.
  2. Multiply all columns Im$ (\sigma_{r+j}(w_i))$ by $ -2i$, thus changing the determinant by $ (-2i)^s$.
  3. Add each columns with entries Re$ (\sigma_{r+j}(w_i))$ to the the column with entries $ -2i$Im$ (\sigma_{r+j}(w_i))$.
Recalling the definition of discriminant, we see that if $ B$ is the matrix constructed by the above three operations, then $ \Det (B)^2 = d_K$. Thus

$\displaystyle \Vol (V/L) = \vert\Det (A)\vert = \vert(-2i)^{-s}\cdot \Det (B)\vert = 2^{-s}\sqrt{\vert d_K\vert}.

$ \qedsymbol$

Lemma 10.4.6   If $ I$ is a nonzero fractional ideal for $ \O _K$, then $ \sigma(I)$ is a lattice in $ V$, and

$\displaystyle \Vol (V/\sigma(I)) = 2^{-s}\sqrt{\vert d_K\vert}\cdot \Norm (I).

Proof. We know that $ [\O _K:I]=\Norm (I)$ is a nonzero rational number. Lemma 10.4.5 implies that $ \sigma(\O _K)$ is a lattice in $ V$, since $ \sigma(\O _K)$ has rank $ n$ as abelian group and spans $ V$, so $ \sigma(I)$ is also a lattice in $ V$. For the volume formula, combine Lemmas 10.4.4-10.4.5 to get

$\displaystyle \Vol (V/\sigma(I)) = \Vol (V/\sigma(\O _K))\cdot[\O _K:I]
=2^{-s}\sqrt{\vert d_K\vert}\Norm (I).

$ \qedsymbol$

Proof. [Proof of Theorem 10.4.2] Let $ K$ be a number field with ring of integers $ \O _K$, let $ \sigma:K\hookrightarrow V\cong \mathbf{R}^n$ be as above, and let $ f:V\to \mathbf{R}$ be the function defined by

$\displaystyle f(x_1,\ldots, x_n) = \vert x_1\cdots x_r\cdot (x_{r+1}^2 + x_{(r+1)+s}^2)\cdots (x_{r+s}^2 + x_n^2).

Notice that if $ x\in K$ then $ f(\sigma(x)) = \vert\Norm _{K/\mathbf{Q}}(x)\vert$.

Let $ S\subset V$ be any closed, bounded, convex, subset that is symmetric with respect to the origin and has positive volume. Since $ S$ is closed and bounded,

$\displaystyle M = \max\{f(x) : x \in S\}


Suppose $ I$ is any nonzero fractional ideal of $ \O _K$. Our goal is to prove there is an integral ideal $ aI$ with small norm. We will do this by finding an appropriate $ a\in I^{-1}$. By Lemma 10.4.6,

$\displaystyle c=\Vol (V/I^{-1}) = \frac{2^{-s}\sqrt{\vert d_K\vert}}{\Norm (I)}.

Let $ \lambda = 2\cdot\left(\frac{c}{v}\right)^{1/n}$, where $ v=\Vol (S)$. Then

$\displaystyle \Vol (\lambda{} S) = \lambda^n \Vol (S) = 2^n \frac{c}{v} \cdot v = 2^n\cdot c=
2^n \Vol (V/I^{-1}),

so by Lemma 10.4.3 there exists $ 0\neq a\in I^{-1}\cap \lambda S$. Since $ M$ is the largest norm of an element of $ S$, the largest norm of an element of $ I^{-1}\cap \lambda{}S$ is at most $ \lambda^n M$, so

$\displaystyle \vert\Norm _{K/\mathbf{Q}}(a)\vert \leq \lambda^n M.

Since $ a\in I^{-1}$, we have $ aI \subset \O _K$, so $ aI$ is an integral ideal of $ \O _K$ that is equivalent to $ I$, and

$\displaystyle \Norm (aI)$ $\displaystyle = \vert\Norm _{K/\mathbf{Q}}(a)\vert\cdot \Norm (I)$    
  $\displaystyle \leq \lambda^n M\cdot \Norm (I)$    
  $\displaystyle \leq 2^n \frac{c}{v} M \cdot \Norm (I)$    
  $\displaystyle \leq 2^n\cdot 2^{-s} \sqrt{\vert d_K\vert} \cdot M \cdot v^{-1}$    
  $\displaystyle = 2^{r+s} \sqrt{\vert d_K\vert} \cdot M \cdot v^{-1}.$    

Notice that the right hand side is independent of $ I$. It depends only on $ r$, $ s$, $ \vert d_K\vert$, and our choice of $ S$. This completes the proof of the theorem, except for the assertion that $ S$ can be chosen to give the claim at the end of the theorem, which we leave as an exercise. $ \qedsymbol$

Corollary 10.4.7   Suppose that $ K\neq \mathbf{Q}$ is a number field. Then $ \vert d_K\vert>1$.

Proof. Applying Theorem 10.4.2 to the unit ideal, we get the bound

$\displaystyle 1\leq \sqrt{\vert d_K\vert}\cdot \left(\frac{4}{\pi}\right)^s\frac{n!}{n^n}.


$\displaystyle \sqrt{\vert d_K\vert}

and the right hand quantity is strictly bigger than $ 1$ for any $ s\leq n/2$ and any $ n>1$ (exercise). $ \qedsymbol$

William Stein 2004-05-06