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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asked Feb 14, 2010 at 15:17
2 Answers
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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.
answered Feb 14, 2010 at 15:57
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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.
answered Feb 14, 2010 at 15:39
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$\begingroup$ 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. $\endgroup$ Commented Feb 14, 2010 at 15:58
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2$\begingroup$ 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. $\endgroup$ Commented Feb 14, 2010 at 16:30
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$\begingroup$ 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. $\endgroup$ Commented Feb 14, 2010 at 16:49 -
10$\begingroup$ 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. $\endgroup$ Commented Feb 14, 2010 at 18:02
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1$\begingroup$ 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. $\endgroup$ Commented Feb 15, 2010 at 6:30
$r_1 + r_2 - 1$independent units. $\endgroup$$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. $\endgroup$