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. (more…)