91Ó°ÊÓ

We need to show that in any finite region of the Euclidean space, there is only a finite number of points that are images of the units in U. Let \(R\) be any finite region in the Euclidean space. Consider an arbitrary point in R and let its coordinates be \((x_1,\dots,x_n)\). Then, there exists constants \(C_1,\dots,C_n\) such that \(C_i \geq |\log|\sigma_i \alpha||\) for all \(\alpha \in U\). This means that \(|\sigma_i(\alpha)| \leq e^{C_i}\) for all \(\alpha \in U\) and \(1 \leq i \leq n\). Now, we claim that there are only finitely many \(\alpha \in U\) satisfying these inequalities. Observe that \(\alpha\) being in U implies that \(\alpha\) is an algebraic integer, and thus \(\alpha\) has a minimal polynomial with integer coefficients. The roots of this polynomial are the conjugates of \(\alpha\), which include the values \(\sigma_i(\alpha)\). Moreover, if all the conjugates of \(\alpha\) have a magnitude less than a certain bound, then the coefficients of the minimal polynomial have bounds that depend on this bound only (by a result from algebraic number theory). Thus, the number of possible minimal polynomials and hence the number of possible \(\alpha\) is finite. In conclusion, for any given finite region R, there can be only a finite number of points in l(U), which means that l(U) is a discrete set in the Euclidean space.

From the previous step, we know that l(U) is a discrete set in the Euclidean space. Since U is a group under multiplication, l(U) is also a group under addition because of the logarithm in l. Moreover, since U is an abelian group, its image l(U) is also abelian. To prove that l(U) is free, we need to find a basis for l(U) and show that it is finitely generated. Let's denote the rank of E over Q by r, meaning that E is a vector space of dimension r over Q. Consider an element \(\alpha \in U\), so the image of the conjugates of \(\alpha\) under the map l is \((\log|\sigma_1(\alpha)|,...,\log|\sigma_n(\alpha)|)\). By the rank-nullity theorem, since the domain of the map l has dimension r, its kernel has dimension at most r. Thus, the image, which is l(U), has dimension at most r in the Euclidean space, meaning that l(U) is finitely generated. Therefore, l(U) is a free abelian group and finitely generated.

The kernel of l consists of all units \(\alpha \in U\) such that l(α) = 0. This means that for all 1 ≤ i ≤ n, we have \(\log|\sigma_i(\alpha)| = 0\). This implies that \(|\sigma_i(\alpha)| = 1\) for all i. Now, consider an element \(\alpha\) in the kernel of l. Since \(|\sigma_i(\alpha)| = 1\), all of its conjugates are complex numbers with magnitude 1, and therefore, they lie on the unit circle in the complex plane. Also, all of the conjugates of \(\alpha\) are algebraic integers, and by the result from algebraic number theory mentioned before, the minimal polynomial of \(\alpha\) must have all of its coefficients be integers. Since the coefficients of the minimal polynomial depend only on the conjugates, this implies that \(\alpha\) is a root of unity, as its minimal polynomial has roots on the unit circle only. Now, we know that the set of roots of unity in any field is finite, as the minimal polynomial of any root of unity can have only finitely many distinct roots. Thus, the kernel of l is a finite group and is the group of roots of unity in E.

We have shown that the image of U under the map l is a free abelian group and finitely generated, and that the kernel of l is a finite group and the group of roots of unity in E. Also, we know U is an abelian group. Since the kernel of l is finite, by the fundamental theorem of abelian groups (the structure theorem for finitely generated abelian groups), U is also a finitely generated abelian group. In conclusion, the group of units U in the ring of algebraic integers OE of the finite extension E over Q is a finitely generated abelian group.