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Put "countable" in the right position.
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Torsten Ekedahl
Torsten Ekedahl
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$\mathbb Q^\ast$ is isomorphic to $\{\pm1\}$ times a countable free abelian group of countable rank. The same is true for an imaginary quadratic field of class number $1$ and different from $\mathbb Q(\sqrt{-3})$ and $\mathbb Q(\sqrt{-1})$ (of which are some though not many). Other examples comes from a real quadratic field of class number $1$ (though this time one of the free factors come from units). Presumably there is an infinite number of those.

$\mathbb Q^\ast$ is isomorphic to $\{\pm1\}$ times a countable free abelian group. The same is true for an imaginary quadratic field of class number $1$ and different from $\mathbb Q(\sqrt{-3})$ and $\mathbb Q(\sqrt{-1})$ (of which are some though not many). Other examples comes from a real quadratic field of class number $1$ (though this time one of the free factors come from units). Presumably there is an infinite number of those.

$\mathbb Q^\ast$ is isomorphic to $\{\pm1\}$ times a free abelian group of countable rank. The same is true for an imaginary quadratic field of class number $1$ and different from $\mathbb Q(\sqrt{-3})$ and $\mathbb Q(\sqrt{-1})$ (of which are some though not many). Other examples comes from a real quadratic field of class number $1$ (though this time one of the free factors come from units). Presumably there is an infinite number of those.

Source Link
Torsten Ekedahl
Torsten Ekedahl
  • 22.6k
  • 2
  • 81
  • 98

$\mathbb Q^\ast$ is isomorphic to $\{\pm1\}$ times a countable free abelian group. The same is true for an imaginary quadratic field of class number $1$ and different from $\mathbb Q(\sqrt{-3})$ and $\mathbb Q(\sqrt{-1})$ (of which are some though not many). Other examples comes from a real quadratic field of class number $1$ (though this time one of the free factors come from units). Presumably there is an infinite number of those.