The Class Group

Definition 7.1.1 (Class Group)   Let $ \O_K$ be the ring of integers of a number field $ K$. The class group $ C_K$ of $ K$ is the group of fractional ideals modulo the sugroup of principal fractional ideals $ (a)$, for $ a\in K$.

Note that if we let $ \Div (\O_K)$ denote the group of fractional ideals, then we have an exact sequence

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

That the class group $ C_K$ is finite 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 (Proposition 6.3.6).

Theorem 7.1.2 (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 Proposition 6.3.6 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. There are other better bounds, but they depend on unproven conjectures.

The following two examples illustrate how to apply Theorem 7.1.2 to compute $ C_K$ in simple cases.

Example 7.1.3   Let $ K=\mathbf{Q}[i]$. Then $ n=2$, $ s=1$, and $ \vert d_K\vert=4$, so the Minkowski bound is

$\displaystyle \sqrt{4} \cdot \left(\frac{4}{\pi}\right)^1 \frac{2!}{2^2}
= \frac{4}{\pi} < 2.

Thus every fractional ideal is equivalent to an ideal of norm $ 1$. Since $ (1)$ is the only ideal of norm $ 1$, every ideal is principal, so $ C_K$ is trivial.

Example 7.1.4   Let $ K=\mathbf{Q}(\sqrt{10})$. We have $ \O_K = \mathbf{Z}[\sqrt{10}]$, so $ n=2$, $ s=0$, $ \vert d_K\vert = 40$, and the Minkowski bound is

$\displaystyle \sqrt{40}\cdot \left(\frac{4}{\pi}\right)^0 \cdot \frac{2!}{2^2}
= 2\cdot \sqrt{10} \cdot \frac{1}{2} = \sqrt{10} = 3.162277\ldots.

We compute the Minkowski bound in SAGE as follows:
sage: K = QQ[sqrt(10)]; K
Number Field in sqrt10 with defining polynomial x^2 - 10
sage: B = K.minkowski_bound(); B
sage: B.n()
Theorem 7.1.2 implies that every ideal class has a representative that is an integral ideal of norm $ 1$$ 2$, or $ 3$. The ideal $ 2\O_K$ is ramified in $ \O_K$, so

$\displaystyle 2\O_K = (2,\sqrt{10}).

If $ (2,\sqrt{10})$ were principal, say $ (\alpha)$, then $ \alpha=a+b\sqrt{10}$ would have norm $ \pm 2$. Then the equation

$\displaystyle x^2 - 10y^2 = \pm 2,$ (7.1)

would have an integer solution. But the squares mod $ 5$ are $ 0,\pm 1$, so (7.1.1) has no solutions. Thus $ (2,\sqrt{10})$ defines a nontrivial element of the class group, and it has order $ 2$ since its square is the principal ideal $ 2\O_K$. Thus $ 2\mid \char93 C_K$.

To find the integral ideals of norm $ 3$, we factor $ x^2-10$ modulo $ 3$, and see that

$\displaystyle 3\O_K = (3,2+\sqrt{10}) \cdot (3,4+\sqrt{10}).

If either of the prime divisors of $ 3\O_K$ were principal, then the equation $ x^2-10y^2 = \pm 3$ would have an integer solution. Since it does not have one mod $ 5$, the prime divisors of $ 3\O_K$ are both nontrivial elements of the class group. Let

$\displaystyle \alpha = \frac{4+\sqrt{10}}{2+\sqrt{10}} = \frac{1}{3}\cdot (1+\sqrt{10}).


$\displaystyle (3,2+\sqrt{10})\cdot (\alpha) = (3\alpha, 4+\sqrt{10})
= (1+\sqrt{10}, 4+\sqrt{10})
= (3, 4+\sqrt{10}),

so the classes over $ 3$ are equal.

In summary, we now know that every element of $ C_K$ is equivalent to one of

$\displaystyle (1),\quad (2,\sqrt{10}),$    or $\displaystyle \quad (3,2+\sqrt{10}).

Thus the class group is a group of order at most $ 3$ that contains an element of order $ 2$. Thus it must have order $ 2$. We verify this in SAGE below, where we also check that $ (3, 2+\sqrt{10})$ generates the class group.
sage: K.<sqrt10> = QQ[sqrt(10)]; K
Number Field in sqrt10 with defining polynomial x^2 - 10
sage: G = K.class_group(); G
Class group of order 2 with structure C2 of Number Field ...
sage: G.0
Fractional ideal class (3, sqrt10 + 1)
sage: G.0^2
Trivial principal fractional ideal class
sage: G.0 == G( (3, 2 + sqrt10) )

Before proving Theorem 7.1.2, we prove a few lemmas. The strategy of the proof is 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$ is an integral ideal, i.e., that $ a\in I^{-1}$.

Let $ S$ be a subset of $ V=\mathbf{R}^n$. Then $ S$ is convex if whenever $ x,y\in S$ then the line connecting $ x$ and $ y$ lies entirely in $ S$. We say that $ S$ is symmetric about the origin if whenever $ x\in S$ then $ -x\in S$ also. If $ L$ is a lattice in the real vector space $ V=\mathbf{R}^n$, then the volume 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 7.1.5 (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. If $ \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$. Because $ S$ is symmetric about the origin, $ -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 nonzero $ Q=Q_\varepsilon \in L\cap {} (1+\varepsilon ) S$ for arbitrarily small $ \varepsilon $. Since $ S$ is closed, $ Q\in L\cap S$. $ \qedsymbol$

Lemma 7.1.6   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\det(A)\vert=[L_1:L_2]$. The claimed formula then follows, since $ [L_1:L_2] = \vert\det(A)\vert$, by definition. $ \qedsymbol$

Fix a number field $ K$ with ring of integers $ \O_K$.

Let $ \sigma_1,\ldots, \sigma_r$ be the real embeddings of $ K$ and $ \sigma_{r+1},\ldots, \sigma_{r+s}$ be half the complex embeddings of $ K$, with one representative of each pair of complex conjugate embeddings. 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$\displaystyle (\sigma_{r+s}(x))\big),$    

Note that this $ \sigma$ is not exactly the same as the one at the beginning of Section 6.2 if $ s>0$.

Lemma 7.1.7  

$\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$\displaystyle (\sigma_{r+s}(w_i)),$   Im$\displaystyle (\sigma_{r+1}(w_i)),\ldots,$   Im$\displaystyle (\sigma_{r+s}(w_i)))

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=\Disc (\O_K)$ in Section 6.2.
  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))$.
  2. Multiply all columns with entries Im$ (\sigma_{r+j}(w_i))$ by $ -2i$, thus changing the determinant by $ (-2i)^s$.
  3. Add each columns that now has entries Re$ (\sigma_{r+j}(w_i))+i$Im$ (\sigma_{r+j}(w_i))$ to the the column with entries $ -2i$Im$ (\sigma_{r+j}(w_i))$ to obtain columns Re$ (\sigma_{r+j}(w_i))-i$Im$ (\sigma_{r+j}(w_i))$.
Recalling the definition of discriminant, we see that if $ B$ is the matrix constructed by doing the above three operations to $ A$, then $ \vert\det(B)^2\vert = \vert d_K\vert$. 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 7.1.8   If $ I$ is a fractional $ \O_K$-ideal, 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. Since $ \sigma(\O_K)$ has rank $ n$ as an abelian group, and Lemma 7.1.7 implies that $ \sigma(\O_K)$ also spans $ V$, it follows that $ \sigma(\O_K)$ is a lattice in $ V$. For some nonzero integer $ m$ we have $ m\O_K \subset I \subset \frac{1}{m}\O_K$, so $ \sigma(I)$ is also a lattice in $ V$. To prove the displayed volume formula, combine Lemmas 7.1.6-7.1.7 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 7.1.2] 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)\vert.

Notice that if $ x\in K$ then $ f(\sigma(x)) = \vert\Norm _{K/\mathbf{Q}}(x)\vert$, and for any $ a\in \mathbf{R}$,

$\displaystyle f(ax_1, \ldots, ax_n) = \vert a\vert^n f(x_1,\ldots, x_n).

Let $ S\subset V$ be a fixed choice of closed, bounded, convex, subset with positive volume 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 fractional ideal of $ \O_K$. Our goal is to prove that there is an integral ideal $ aI$ with small norm. We will do this by finding an appropriate $ a\in I^{-1}$. By Lemma 7.1.8,

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

so by Lemma 7.1.5 there exists $ 0\neq b\in \sigma(I^{-1})\cap \lambda S$. Let $ a\in I^{-1}$ be such that $ \sigma(a)=b$. Since $ M$ is the largest norm of an element of $ S$, the largest norm of an element of $ \sigma(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

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

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

Proof. Applying Theorem 7.1.2 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$

A prime $ p$ ramifies in $ \O_K$ if and only if $ d\mid d_K$, so the corollary implies that every nontrivial extension of $ \mathbf {Q}$ is ramified at some prime.

William Stein 2012-09-24