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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?

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Nice question. I have thinking about something somewhat related lately: every f.g. domain has a f.g. unit group. Everywhere I look, the proof uses the Dirichlet unit theorem. But it doesn't feel like that should be necessary. – Steve D Feb 14 '10 at 15:29
I guess that what you're expecting is a (mildly) simple formula for a unit of infinite order. Even in the easiest (quadratic) case, AFAIK there is no closed-form formula in terms of d>0 for x(d),y(d) such that x(d)^2-dy(d)^2=1 (and y(d) nonzero), so it seems rather unlikely. – Ewan Delanoy Feb 14 '10 at 15:50
@Ewan: I realized that there probably couldn't be a simple formula because of the Pell's equation case you mentioned, but still thought that there might be a simpler proof, since the unit theorem proves existence of $r_1 + r_2 - 1$ independent units. – Victor Miller Feb 14 '10 at 15:53
I've been thinking about the case of real abelian fields. In the case of the real subfield of $\mathbb{Q}(\zeta_n)$ we do have an explicit formula for the cyclotomic units, and the action of the Galois group on them is completely explicit. So to show the existence of a non-trivial unit in a subfield we just need to show that the things fixed by a the corresponding subgroup of galois is non-trivial. This would seem to be a linear algebra problem over finite $\mathbb{Z}$-modules. – Victor Miller Feb 14 '10 at 17:56
Since it took me longer than I would have liked to come up with a proof of the fact that if $L/K$ is not CM and not trivial then the unit rank $r(L)$ is greater than $r(K)$, here it is (for the edification of others): $d=\deg(L/K)$, primes refer to $L$, non-primes to $K$: $r_1' + 2r_2'=d(r_1 + 2r_2)$. Let $s = r_2' - d r_2 \ge 0$. If $r_2' - r_2 + r_1' - r_1 \le 0$` then $(d-1) (r_1 + r_2) \le s \le (d/2) r_1$. If $d > 2$ this implies $r_1 = r_2 = 0$ which is impossible. If $d=2$ this implies `$r_2 = r_1' = 0$ which is CM. – Victor Miller Feb 14 '10 at 23:18

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.

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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.

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Franz, thanks. I wonder if it's possible to give a good characterization of those fields that do have such a generator. It looks like the most likely exceptions are abelian fields but things may be more subtle than that. – Victor Miller Feb 14 '10 at 15:58
I've corrected my initial claims. Hunter has given a counterexample to your question below which can be generalized to most CM fields (totally complex quadratic extensions of totally real fields). For the rest, your question is still open. – Franz Lemmermeyer Feb 14 '10 at 16:30
Franz, so does the argument go something like this: Let $K=\mathbb{Q}(\zeta_n)$ chosen so that the index of of unit group in the real subfield of index 2 in the group of units is 1 (or maybe prime to $2n$). Choose $H$ a proper subgroup of the galois group not containing complex conjugation and index $> 2$, and take the fixed field by $H$. The idea is to kill off the roots of unity, as in Hunter's answer. – Victor Miller Feb 14 '10 at 16:49
If K is not a CM field, then K is generated by a unit: K has only finitely many subfields, each of which has lower unit rank than K, so it is possible to choose a unit of K outside these subfield unit groups. If K is a CM field, sometimes K is generated by a unit and sometimes not, but in any case there exists m>0 such that the m-th power of every unit u is in the totally real subfield. – Bjorn Poonen Feb 14 '10 at 18:02
The quantity to look at in the CM-case L/K (L totally complex, K totally real, (L:K) = 2) is Hasse's unit index Q=(E_L:W_L E_K), where E_* denotes the unit group and W_* the group of roots of unity in the fields. This index is 1 or 2, and can be computed for quite general families of fields (see Acta Arith. 72 (1995), 347-359 for a collection of known results). If Q = 1, obstructions to the existence of a "generating unit" must come from the roots of unity. If Q = 2, the problem seems to lie deeper, but in any case there is a unit not contained in the real subfield K. – Franz Lemmermeyer Feb 15 '10 at 6:30

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