Two questions about units in Number Fields If $K/\mathbb{Q}$ is a number field which is not $\mathbb{Q}$ or a quadratic imaginary field then, by the Dirichlet unit theorem it has a unit of infinite order.  Is there a simple proof of this fact which doesn't haul in all the machinery of the unit theorem?  In a related question is it true that such a $K$ is always generated over $\mathbb{Q}$ by one unit (which might not be of infinite order -- cf. the case of a full cyclotomic field), and if it is, is there a simple proof of this fact?
 A: Dirichlet's unit theorem is a relatively straightforward extension of the well known proof that the Pell equation has a nontrivial solution by a clever use of Dirichlet's box principle. The "machinery" is only needed to control the combinatorial explosion, or,
as far as geometry of numbers is concerned, as a natural generalization of the box principle.
As for your second question I first thought that the answer is no. Take for example a biquadratic number field $K = {\mathbb Q}(\sqrt{m},\sqrt{n}\,)$, with $m,n > 0$ chosen in such a way that the unit index (units of K : units from the subfields) is 1. This implies that the only units in K are multiples of those coming from the three subfields. But you still get generators of the field by taking the product of two units coming from different subfields, so your question is still open.
A: The answer to question 2 as stated is no. Let $K = \mathbb{Q}(\sqrt{p},\sqrt{-b})$, with $p$ 1 mod 4. By Frohlich and Taylor, p. 196, the fundamental unit group is generated by a unit from the real quadratic subfield. It then suffices to show that we can pick b so that K contains no roots of unity other than $\pm 1$. But the only possibility is that K contains a 3rd, 4th, or 5th root of unity, and so we just pick $b$ to avoid ramification at 2, 3, or 5.
