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Next: 6.7 Prime ideals Up: 6 Algebraic Number Fields Previous: 6.5 Groups of invertible

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 TM = $ \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 TM with the property that vol(TM) = vol(TR)Nm(M). Now, if A is any (compact measurable) subset of $ \mathbb {R}$ . K with the property that vol(A) > vol(TM) then the map A$ \to$TM cannot be one-to-one (with a little thought it is clear that this is actually also true if vol(A) $ \geq$ vol(TM)). 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.

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

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}$wn, then the volume of TR, with respect to this identification is the determinant of the n×n matrix $ \Omega$ given by

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

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

$\displaystyle \tilde{\Omega}$ = $\displaystyle \left(\vphantom{
}\right.$wi(1),..., wi(r1),$\displaystyle \tilde{w}_{i}^{(1)}$,$\displaystyle \overline{\tilde{w}_i^{(1)}}$,...,$\displaystyle \tilde{w}_{i}^{(r_2)}$,$\displaystyle \overline{\tilde{w}_i^{(r_2)}}$$\displaystyle \left.\vphantom{

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

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

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

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

vol(A) = 2r1$\displaystyle \pi^{r_2}_{}$$\displaystyle \prod_{i=1}^{r_1}$ai$\displaystyle \prod_{j=1}^{r_2}$bj2

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

2r1$\displaystyle \pi^{r_2}_{}$$\displaystyle \prod_{i=1}^{r_1}$ai$\displaystyle \prod_{j=1}^{r_2}$bj2 = (1/2r2)$\displaystyle \sqrt{\vert D_R\vert}$Nm(M)

Now the norm of the element v is the product

Nm(v) = $\displaystyle \prod_{i=1}^{r_1}$| v1(i) - v2(i)| . $\displaystyle \prod_{j=1}^{r_2}$|$\displaystyle \tilde{v}_{1}^{(j)}$ - $\displaystyle \tilde{v}_{2}^{(j)}$|2 $\displaystyle \leq$ 2r1$\displaystyle \prod_{i=1}^{r_1}$ai×22r2$\displaystyle \prod_{j=1}^{r_2}$bj2

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) $\displaystyle \leq$ % latex2html id marker 16411
$\displaystyle \left(\vphantom{\frac{2}{\pi}}\right.$% latex2html id marker 16412
$\displaystyle {\frac{2}{\pi}}$% latex2html id marker 16413
$\displaystyle \left.\vphantom{\frac{2}{\pi}}\right)^{r_2}_{}$$\displaystyle \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!

next up previous
Next: 6.7 Prime ideals Up: 6 Algebraic Number Fields Previous: 6.5 Groups of invertible
Kapil Hari Paranjape 2002-10-20