There is another major result in algebraic number theory that we need to get to!  I have this no longer secret goal of getting to class field theory, and if it happens, this will be a key result.  The hard part of the actual proof (namely, the determination of the rank of a certain lattice) will be deferred until next time; it’s possible to do it with the tools we already have, but it is cleaner (I think) to do it once ideles have been introduced.

Following the philosophy of examples first, let us motivate things with an example. Consider the ring {\mathbb{Z}[i]} of Gaussian; as is well-known, this is the ring of integers in the quadratic field {\mathbb{Q}(i)}. To see this, suppose {a+ib, a, b \in \mathbb{Q}} is integral; then so is {a-ib}, and thus {2a,2b} are integers. Also the fact {(a+ib)(a-ib) = a^2+b^2 } must be an integer now means that neither {a,b} can be of the form {k/2} for {k} odd.

What are the units in {\mathbb{Z}[i]}? If {x} is a unit, so is {\bar{x}}, so the norm {N(x)} must be a unit in {\mathbb{Z}[i]} (and hence in {\mathbb{Z}}). So if {x=a+ib}, then {a^2+b^2=1} and {x = \pm 1} or {\pm i}. So, the units are just the roots of unity.

In general, however, the situation is more complicated. Consider {\mathbb{Z}[\sqrt{2}]}, which is again integrally closed. Then {x=a+b\sqrt{2}} is a unit if and only if its norm to {\mathbb{Q}}, i.e. {a^2 - 2b^2} is equal to {\pm 1}. Indeed, the norm {N(x)} of a unit {x} is still a unit, and since {\mathbb{Z}} is integrally closed, we find that {N(x)} is a {\mathbb{Z}}-unit. In particular, the units correspond to the solutions to the Pell equation. There are infinitely many of them.

But the situation is not hopeless. We will show that in any number field, the unit group is a direct sum of copies of {\mathbb{Z}} and the roots of unity. We will also determine the rank.

1. {S}-units

Let {K} be a number field. Then I claim that {x \in K} is a unit if and only if its order at any nonarchimedean valuation is zero. Indeed, if we let {\mathcal{O}} be the ring of integers in {K}—so that {\mathcal{O}} is a Dedekind domain—then {x} belongs to the localizations {\mathcal{O}_{\mathfrak{p}}} for each nonzero prime ideal {\mathfrak{p}}. But

\displaystyle \mathcal{O} = \bigcap_{\mathfrak{p}} \mathcal{O}_{\mathfrak{p}}

from a basic result in commutative algebra true of any integral domain. Hence if {x} satisfies the order condition, then {x \in \mathcal{O}}, and by symmetry {x^{-1} \in \mathcal{O}}, proving the claim.

Now, let {S} be a finite set of places of {K} containing the archimedean places {S_\infty}. We say that {x \in K} is a {S}-unit if {x} has order zero at the places outside {S}; denote the set of {S}-units by {K_S}.

This is clearly a generalization of the usual notion of unit, and reduces to it when {S=S_{\infty}}. Alternatively, if we consider the ring {\mathcal{O}_S} of elements of {K} integral outside {S}, then {K_S} is the unit group of {\mathcal{O}_S}.

2. The unit theorem

The principal goal in this post is to prove:

Theorem 1 (Unit theorem) The group {K_S} is isomorphic to a direct sum {\mathbb{Z}^{|S|-1} \bigoplus G}, where {G} is the group of roots of unity in {K}.


To prove it, we define a mapping {h: K_S \rightarrow \mathbb{R}^{S}} by (where {N_v = [K_v: \mathbb{Q}_{v_0}]})

\displaystyle h(x) = ( N_v \log |x|_v )_{v \in S}

so that by the product formula, {h(K_S)} is contained in the subspace {W \subset \mathbb{R}^S} defined by {(x_v)_{v \in S} \in W} iff {\sum x_v = 0}.

I claim that: first, the kernel of {h} consists of the roots of unity; and second, the image of {h} is a lattice in {W}. Together, these will prove the unit theorem after we have computed the rank (which we will do later).

Claim 1. If {x \in \mathrm{ker} \ h}, then {|x|_v = 1} for all places {v}. We now need a lemma.

Lemma 2 Suppose {C} is a constant. Then there are only finitely many points {x} of {K} such that {|x|_v \leq C} for all places {v}.

The characteristic polynomial of {x} (as an endomorphism of {K}) has rational coefficients which are the symmetric functions of the conjugates of {x}, and which thus are bounded at all rational places by {D=D(C)}. However, there are only finitely many rational numbers {r} satisfying {|r|_p \leq D(C), \ \forall p} and { |r|_{\infty} \leq D(C)}, as is easily checked, so there are only finitely many possible characteristic polynomials. Now {x} satisfies its characteristic polynomial, so there are only finitely many possibilities for it. This proves the lemma.

To finish the proof of the claim, note that all powers of {x} have norm 1 at all places, so by the lemma we must get {x^m = x^n} for some {m \neq n}, proving the claim.

Note in particular that {h(K_S)} must be a discrete set by the lemma, since any bounded region in {\mathbb{R}^S} contains only finitely many points. Thus {h(K_S)} is a lattice in {\mathbb{R}^S}, and {K_S} is the direct sum of a finite group (the roots of unity) and a lattice.

It remains to determine the rank of the lattice, which we will do next time; it will be easier to do this after ideles have been introduced.