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What are all ordered pairs $(n,m)$ such that the multiplicative groups of the fields $\mathbb{Q}(\sqrt{n})$ and $\mathbb{Q}(\sqrt{m})$ are isomorphic?

I saw a question earlier today claiming that $(3,7)$ is one such pair (although I'm not sure this is true), and I'm interested in generalizing this.

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@Joseph: the claim isn't that they're isomorphic in a way that preserves $\mathbb{Q}$. –  Qiaochu Yuan Jun 14 '13 at 0:09

2 Answers 2

up vote 4 down vote accepted

The multiplicative group $\Bbb Q(\sqrt d)^*$ is abstractly isomorphic to $U \times \bigoplus_\omega \Bbb Z$, where $U$ is the torsion subgroup of $\Bbb Q(\sqrt d)^*$, which means it is the roots of unity in $\Bbb Q(\sqrt d)^*$.

$\Bbb Q(\sqrt{-1})^*$ and $\Bbb Q(\sqrt{-3})^*$ both have $U$ larger than $\{-1;+1\}$, so those two are special cases and are not isomorphic to anyone else but themselves. For every other quadratic field, $U = \{-1;+1\}$, and so those $\Bbb Q(\sqrt d)^*$ are all abstractly isomorphic to each other.

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For example, with $m,n<0$, if in both cases the rings of integers are PIDs, and if the fields contain no roots of unity beyond $\pm 1$, then the multiplicative groups are free groups on primes-mod-units, with $\pm 1$ thrown in. That is, in some cases the answer can be "yes, but for boring reasons". For $m,n>0$, the units group mod $\pm 1$ in the ring of algebraic integers is $\mathbb Z$, by Dirichlet, with no differentiation depending on $m,n$.

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If the field is not a PID, then number field elements-mod-units is still a subgroup of the free group generated by the prime ideals, and subgroups of free groups are still free. And the group of powers of a prime is isomorphic to the group of powers of a unit. –  Hurkyl Jun 14 '13 at 9:07

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