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|*w*1(*u*)|, log|*w*2(*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 *x*1 satisfying the conditions above. Then replace one of the *n* points by *x*1, 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**.

[…] From the previous math post, the map from the unit group of a number field K of degree n over Q to the vector space R^n has an image with the structure of Z^q where q <= n. In fact, we can say more: if K has r real conjugates and 2s complex ones, then since the map sends the unit group to a subspace of R^n of dimension r + s – 1, we can say that q <= r + s – 1; this is because Z^q is a lattice. […]