6.6 Minkowski's Geometry of Numbers

In order to decide whether or not a lattice M is of the form $\alpha$ ^. R (and hence trivial in the class group) we need to find elements $\alpha$ in M whose norm is as close as possible to that of M. This is achieved in the following section.

We now want to give a ``measure'' associated with an order R. The space of n×n matrices with rational entries is naturally contained in the space of n×n matrices with real entries. Thus we can consider the ring $\mathbb {R}$ ^. K of real linear combinations of elements of K. This is an n-dimensional vector space over $\mathbb {R}$. Thus, for any lattice M, the space T_M = $\mathbb {R}$ ^. K/M is an n-dimensional torus. Taking some translation invariant measure on $\mathbb {R}$ ^. K gives us a notion of volume for the tori T_M with the property that vol(T_M) = vol(T_R)Nm(M). Now, if A is any (compact measurable) subset of $\mathbb {R}$ ^. K with the property that vol(A) > vol(T_M) then the map A$\to$T_M cannot be one-to-one (with a little thought it is clear that this is actually also true if vol(A) $\geq$ vol(T_M)). The difference between two points with the same inverse image will give a non-zero element of M.

Now, one natural way to identify $\mathbb {R}$ ^. K with $\mathbb {R}$ⁿ (and thus put a measure on it) is to use ``simultaneous diagonalisation''. As seen above K is generated by a single n×n matrix $\alpha$ whose characteristic polynomial P(T) is irreducible over rationals. This means that this has distinct roots and so over real numbers can be brought into a ``diagonal'' form as below by a suitable change of co-ordinates.

$$\begin{pmatrix} \alpha^{(1)} & 0 & \dots & 0 & 0 & 0 & \dots &0 ... ...de{\alpha}^{(r_2)} & \operatorname{\rm Re}\tilde{\alpha}^{(r_2)} \end{pmatrix}$$

Here $\alpha^{(i)}_{}$ denote the real roots of P(T), while ($\tilde{\alpha}^{(j)}_{}$,$\overline{\tilde{\alpha}^{(j)}}$) are the pairs of conjugate complex roots of P(T). Now, every element of K is a linear combination of powers of $\alpha$ so that it too is brought into the above form by the same change of co-ordinates. For simplicity of notation we write the matrix associated with an element $\beta$ of K as [$\beta^{(1)}_{}$,...,$\beta^{(r_1)}_{}$,$\tilde{\beta}^{(1)}_{}$,...,$\tilde{\beta}^{(r_2)}_{}$]. More generally, for any element x in $\mathbb {R}$ ^. K we have a representation [x⁽¹⁾,..., x^(r₁),$\tilde{x}^{(1)}_{}$,...,$\tilde{x}^{(r_2)}_{}$]. This representation gives us an identification of $\mathbb {R}$ ^. K with $\mathbb {R}$ⁿ. If R = $\mathbb {Z}$ ^. w₁ + ... + $\mathbb {Z}$w_n, then the volume of T_R, with respect to this identification is the determinant of the n×n matrix $\Omega$ given by

$$\Omega$$ = $$\left(\vphantom{ w_i^{(1)},\dots,w_i^{(r_1)},\operatorname{\rm R... ...e{\rm Re}\tilde{w_i}^{(r_2)},\operatorname{\rm Im}\tilde{w_i}^{(r_2)} }\right.$$w_i⁽¹⁾,..., w_i^(r₁), Re$$\tilde{w_i}^{(1)}_{}$$, Im$$\tilde{w_i}^{(1)}_{}$$,..., Re$$\tilde{w_i}^{(r_2)}_{}$$, Im$$\tilde{w_i}^{(r_2)}_{}$$$$\left.\vphantom{ w_i^{(1)},\dots,w_i^{(r_1)},\operatorname{\rm R... ...tilde{w_i}^{(r_2)},\operatorname{\rm Im}\tilde{w_i}^{(r_2)} }\right)_{i=1}^{n}$$

Let the matrix $\tilde{\Omega}$ (complex entries) be given by

$$\tilde{\Omega}$$ = $$\left(\vphantom{ w_i^{(1)},\dots,w_i^{(r_1)},\tilde{w}_i^{(1)},\... ...}_i^{(1)}}, \dots,\tilde{w}_i^{(r_2)},\overline{\tilde{w}_i^{(r_2)}} }\right.$$w_i⁽¹⁾,..., w_i^(r₁),$$\tilde{w}_{i}^{(1)}$$,$$\overline{\tilde{w}_i^{(1)}}$$,...,$$\tilde{w}_{i}^{(r_2)}$$,$$\overline{\tilde{w}_i^{(r_2)}}$$$$\left.\vphantom{ w_i^{(1)},\dots,w_i^{(r_1)},\tilde{w}_i^{(1)},\... ..., \dots,\tilde{w}_i^{(r_2)},\overline{\tilde{w}_i^{(r_2)}} }\right)_{i=1}^{n}$$

Standard rules for column operations on determinants show that the determinant of $\tilde{\Omega}$ is 2^r₂ times the determinant of $\Omega$. On the other hand the (i, j)-th entry of the matrix $\tilde{\Omega}$ ^. $\tilde{\Omega}^{t}_{}$ is

$$\sum_{p=1}^{r_1}$$w_i^(p)w_j^(p) + $$\sum_{q=1}^{r_2}$$$$\tilde{w}_{i}^{(q)}$$$$\tilde{w}_{j}^{(q)}$$ + $$\sum_{q=1}^{r_2}$$$$\overline{\tilde{w}_i^{(q)}}$$ $$\overline{\tilde{w}_j^{(q)}}$$

which we immediately recognise as Trace(w_i ^. w_j) when it is expressed in the form given above. Combining these observations we obtain the identity vol(T_R) = (1/2^r₂)$\sqrt{\vert D_R\vert}$.

Now consider the region A consisting of all x in $\mathbb {R}$ ^. K so that | x⁽ⁱ⁾| $\leq$ a_i and |$\tilde{x}^{j}_{}$| $\leq$ b_i for some positive constants a_i and b_j. We have

vol(A) = 2^r₁$$\pi^{r_2}_{}$$$$\prod_{i=1}^{r_1}$$a_i$$\prod_{j=1}^{r_2}$$b_j²

Thus, in order to obtain a pair (v₁, v₂) in A so that v = v₁ - v₂ is a non-zero element of M we need the condition

2^r₁$$\pi^{r_2}_{}$$$$\prod_{i=1}^{r_1}$$a_i$$\prod_{j=1}^{r_2}$$b_j² = (1/2^r₂)$$\sqrt{\vert D_R\vert}$$Nm(M)

Now the norm of the element v is the product

Nm(v) = $$\prod_{i=1}^{r_1}$$| v₁⁽ⁱ⁾ - v₂⁽ⁱ⁾| ^. $$\prod_{j=1}^{r_2}$$|$$\tilde{v}_{1}^{(j)}$$ - $$\tilde{v}_{2}^{(j)}$$|² $$\leq$$ 2^r₁$$\prod_{i=1}^{r_1}$$a_i×2^2r₂$$\prod_{j=1}^{r_2}$$b_j²

Hence, we have the following

Lemma 14   For any ideal I of an order R in K there is a non-zero element v in I so that

Nm(v) $$\leq$$ $$\left(\vphantom{\frac{2}{\pi}}\right.$$$${\frac{2}{\pi}}$$$$\left.\vphantom{\frac{2}{\pi}}\right)^{r_2}_{}$$$$\sqrt{\vert D_R\vert}$$ ^. | R/I|

Here we have written | R/I| instead of Nm(I) in order to make the dependence on R clear.

Proof. We just combine the inequality above with the condition that needs to be satisfied in order to obtain such a v. $\qedsymbol$

In particular, if I is an invertible ideal, then vR = I ^. J where J = vR ^. I^-1 and Nm(J) $\leq$ (2/$\pi$)^r₂$\sqrt{\vert D_R\vert}$. Now, for any element of the class group Cl(R), let I represent the inverse of this class. The above argument produces a representative J of the class which has norm no more than (2/$\pi$)^r₂$\sqrt{\vert D_R\vert}$. In particular, we have shown than the class group is finite (an ideal J of norm n is a quotient of cardinality n the group R/nR; there are at most finitely many such quotient groups).

While it is not too difficult to use this procedure to write all the ideals J satisfying the above condition, it is much harder to write the ``multiplication table'' for the group Cl(R) on the basis of what has gone so far. If J₁ and J₂ are two ideals as above and the product no longer satisfies the above condition, then we need to find the element v in (J₁ ^. J₂) that the lemma guarantees. But the proof of the lemma gives us no way to find such elements!