Let M be euclidean space, i.e. real space, with r1+r2 dimensions. Apply a continuous map from complex space into M by taking the log of the absolute value of each component. Conjugate pairs have the same norm, so we just add these together to make one real value. This is not defined on the hyperplanes where any of the components are equal to 0. Remember that a conjugate of x is 0 iff all conjugates are 0, iff x = 0, so this restriction only excludes the point 0 from S. Everything else maps into complex space, and into M. I will call this the "log" embedding of S into euclidean space. It should not be confused with the component embedding used in the geometry of numbers.
The log of the absolute value is a continuous function, hence the map into M is continuous per component, and continuous wherever it is defined.
Let U be the group of units in S. Group action is multiplication in S as usual. Multiplication commutes with each embedding, hence the product in S becomes the product in each coordinate of complex space. Multiplication also commutes with absolute value. After taking logs, multiplication becomes addition in each coordinate of M. Thus we have a group homomorphism from U into M, where multiplication in U becomes addition in M.
Let K be the kernel, the units that map to 0. Thus units in K have absolute values of 1 over all embeddings. Define a bounded shape in complex space, with components bounded by 1. There are finitely many lattice points in this set, which pull back to finitely many points of S. Therefore K is finite.
A multiplicative finite group in a field is cyclic, hence K is cyclic, generated by some rth root of 1. Conversely, every root of 1 in E is integral over Z, and lies in S. Therefore K consists of the roots of 1 in E, a finite cyclic group.
Note that K contains all torsion elements of U. This because a torsion element in a multiplicative group is always a root of 1. Therefore the quotient U/K, which is the image of U, is torsion free.
The image of U is closed under addition in M. If 0 is not a cluster point, we have a lattice. Place a small closed ball about 0 and pull back, by continuity, to a closed bounded region in complex space. This contains finitely many lattice points. Thus there are finitely many points of S, and of U, near 0 in M, and U becomes a lattice in M.
This is a Z lattice, or a free Z module, whose rank is bounded by r1+r2, the dimension of M. In fact, the rank is one less. The norm of a unit, times the norm of its inverse, equals the norm of 1, which is 1. Therefore the norm of any unit in S is a unit in Z, which is ±1. The norm is the product of the conjugates. Thus the product of the conjugates in complex space becomes ±1. Map this into M by taking logs, and the sum of the real logs, plus twice the sum of the complex logs, equals log(1), which is 0. The image of U lies in an r1+r2-1 dimensional hyperplane passing through M. Specifically, the coordinates of M must sum to 0.
In summary, U is finitely generated, with a cyclic torsion subgroup consisting of the roots of 1 in E, and a free subgroup with rank ≤ r1+r2-1.
For a negative quadratic extension, r1+r2-1 = 0, and the units are just the roots of 1 in E. These are ±1, ±i if sqrt(-1) is adjoined, and the sixth roots of 1 for Z adjoin (1+sqrt(-3))/2. See quadratic number fields for more details.