In the previous post, we considered the map from O(K)*, the group of units of the ring of integers of the number field K, to the additive group of the vector space R^n, where n is [K:Q], the degree of K over Q. The map u –> (log|w1(u)|, log|w2(u)|, …, log|w(n)(u)|) is a group homomorphism from the multiplicative group O(K)* to the additive group R^n, such that u –> 0 iff u^k = 1 for some k.
In this post, I’m going to prove that the image of the map is a lattice in R^n, i.e. it has the structure of Z^q for some q <= n. But in the previous post, I proved that for every real number c, there are only finitely many units u such that |w(i)(u)| <= c for all i. That means that for every c, there are only finitely many units u such that log|w(i)(u)| <= log(c) for every i. By definition, every bounded set S in R^n has some c such that the coordinates of every member of S are all at most c, so the image of the map intersects every bounded set in a finite set.
The definition I gave of a lattice is the additive group generated by linearly independent elements. An equivalent definition is an additive group that intersects every bounded set in a finite set. To see why, every lattice that has the structure of Z^m, where m <= n, clearly intersects a set bounded by c in a set with at most (c+1)^m points. It’s the converse that requires work.
First, assume that no point in R^n is linearly independent of the additive subgroup G over R. If it doesn’t hold, just pass to the R-linear span of G, which will be R^m for some m < n.
Now, take n different R-linearly independent points of G; these form an R-basis for R^n, so we can write every point of R^n in terms of them. If except the origin there exists no point x of G that can be written in terms of these n points such that every coordinate of x is between 0 and 1 including 0 but not 1, then these points generate G over Z. To see why, if a point’s coordinates in terms of these n points are not all integers, then we can subtract the integer parts to get another point of G that is not zero and has every coordinate between 0 and 1 including 0 but not 1. In that case, G would be generated by linearly independent points, so it’d be a lattice.
Now, suppose that there does exist an x1 satisfying the conditions above. Then replace one of the n points by x1, and repeat the process. Every point x(i) will be bounded by 1 in terms of the original n points; and it’s possibly to choose the point to replace with x(i) in such a way that the volume of the region of points with the required coordinates will decrease. This ensures that no two x(i)’s are equal. If G intersects every bounded set in a finite set, then there are only finitely many such x(i)’s, so eventually we’ll get to a set of n points that generate G over Z.