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

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

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

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

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

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

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

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

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

$$\sigma(x) = \big( \sigma_1(x), \sigma_2(x),\ldots, \sigma_r(x),$$

$$(\sigma_{r+1}(x)), \ldots, (\sigma_{r+s}(x)), (\sigma_{r+1}(x)), \ldots, (\sigma_{r+s}(x))\big),$$

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  

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

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

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

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

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

$\qedsymbol$

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

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

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

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

$\qedsymbol$

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

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

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,

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

exists.

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,

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

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

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

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

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

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

$$\Norm (aI) = \vert\Norm _{K/\mathbf{Q}}(a)\vert\cdot \Norm (I)$$

$$\leq \lambda^n M\cdot \Norm (I)$$

$$\leq 2^n \frac{c}{v} M \cdot \Norm (I)$$

$$\leq 2^n\cdot 2^{-s} \sqrt{\vert d_K\vert} \cdot M \cdot v^{-1}$$

$$= 2^{r+s} \sqrt{\vert d_K\vert} \cdot M \cdot v^{-1}.$$

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

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

Thus

$$\sqrt{\vert d_K\vert} \geq \left(\frac{\pi}{4}\right)^s\frac{n^n}{n!},$$

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