Someone hit my blog by Googling *Multiplicative group of Q(sqrt(2))*. The post s/he read, the one about roots of unity, only talks about the multiplicative group of roots of unity. The short answer to the question is that **Q**(SQRT(2)) has the same multiplicative structure as **Q**, i.e. the multiplicative groups are isomorphic.

I also found this discussion about isomorphisms of multiplicative groups, where people erroneously believe that **Q**(SQRT(2)) and **Q**(SQRT(3)) don’t have isomorphic multiplicative groups. In fact they do, although the isomorphism is very different from the isomorphism of the additive groups.

Now, on to the long answer. For any set S, we can form the group **Z**^S, comprised of all functions from S to **Z** such that all but finitely many elements of S map to 0. The group operation is pointwise addition of functions, the identity is the everywhere zero function, and the inverse of *f* is the function that sends each *s* to –*f*(*s*). Associativity and commutativity are obvious, so this is an abelian group.

Axiomatic set theorists represent 0 as the empty set, 1 as the set {0}, 2 as {0, 1}, 3 as {0, 1, 2}, etc., so that the set *n* has *n* elements. So, for example, **Z**^2 is the set of all functions from {0, 1} to **Z**, where each function can be represented as (*f*(0), *f*(1)), which is the usual way to denote elements of **Z**^2.

Groups of this form are called free abelian, because they always have an integral basis, comprised of functions of the form *f*(*x*) = 1 if *x* = *s* and 0 otherwise, where *s* is any element of S. The integral basis is free, i.e. if *a*1*f*1 + *a*2*f*2 + … + *a*(*n*)*f*(*n*) = 0 where the *f*(*i*)’s are distinct basis elements and the *a*(*i*)’s are integers, then *a*(*i*) = 0 for all *i*.

Any function *h* from S to T induces a group homomorphism from **Z**^S to **Z**^T, defined by *g*(*f*(*s*)) = *f*(*h*(*s*)). It’s not difficult to see that *g* is injective iff *h* is, and is surjective iff *h* is. In other words, free abelian groups are completely classified up to isomorphism by the sizes of their underlying sets, where we say S <= T if there’s an injective map from S to T, S = T if S <= T and T <= S, and S < T if S <= T and T !<= S.

Also, suppose that A is an abelian group, **Z**^S is free abelian, and *g* is a surjective group homomorphism from A to **Z**^S. Since **Z**^S has an integral basis, every group homomorphism from it to another group is uniquely determined by the images of the integral basis elements. So define *g*‘ from **Z**^S to A by letting *g*‘(*f*(*i*)) be any element mapping to *f*(*i*) under *g*. The *f*(*i*)’s form an integral basis, so any choice will do. Then it’s trivial to show that *g*(*g*‘(*f*)) = *f* for all *f*. Not all abelian groups have this property, which is called splitting – for example, this trick doesn’t work with the map from **Z** to **Z**/2**Z**.

Now, back to number fields. The ring of integers of a number field is a Dedekind domain, so it has unique factorization into ideals. For now, let’s assume that we’re also talking about PIDs, a harmless assumption in the case of K = **Q**(SQRT(2)) or K = **Q**(SQRT(3)). The group of fractional ideals of K is free abelian on the set of prime ideals, because any fractional ideal I can be represented as a function mapping each prime ideal to the degree to which it divides I.

The group homomorphism from K* to F(K) that sends *a* to (*a*) is surjective, since K is a PID. So we can use the splitting property of free abelian groups to lift each prime ideal to some element of K. If the homomorphism were also injective then it’d show that K* = F(K), but the homomorphism is in fact never injective, since it maps -1 to (1). Still, all that remains is to deal with the kernel, which is the group of units of R.

But now the group of units is W * **Z**^(*r* + *s* – 1), i.e. W times a free abelian group. The product of **Z**^S and **Z**^T is free abelian on the disjoint union of S and T, so K* is just W times a free abelian group, whose underlying set is the set of all prime ideals of K plus *r* + *s* – 1 linearly independent units.

But the underlying set is countable, i.e. in bijection with **N**. To see why, we can order all elements in the underlying set as a sequence: first, we can have the linearly independent units. Then, we can have the at most *n* prime ideals lying over 2, then the prime ideals lying over 3, and so on. This sequence is infinite, since every prime integer has at least one prime lying over it, and there are infinitely many primes. It also hits every member of the underlying set, so it induces a bijection from the set to **N** (it can’t be done for every set; however, K = **N** as sets for every number field K, and even **A** = **N**).

We can do that for any number field with class number 1, so all number fields with class number 1 and the same group of roots of unity have isomorphic multiplicative groups. In particular, this applies to all number fields with real conjugates, since in that case the group of roots of unity is necessarily just {-1, 1}. So multiplicatively, **Q** = **Q**(SQRT(2)) = **Q**(SQRT(3)) = **Q**(SQRT(-2)) = …

If you want an explicit three-way isomorphism, then let’s first concentrate on the group of positive elements of **Q**/**Q**(SQRT(2))/**Q**(SQRT(3)) and then multiply by {-1, 1}. A natural order for the underlying set of **Q** is (2, 3, 5, 7, 11, 13, …); a natural order for this of **Q**(SQRT(2)) is (1 + SQRT(2), SQRT(2), 3, 5, 3 + 2SQRT(2), 3 – 2SQRT(2), …); and a natural order for this of **Q**(SQRT(3)) is (2 + SQRT(3), 1 + SQRT(3), -1 + SQRT(3), SQRT(3), 5, 7, …).