TITLE: How to show that $(\mathbb{Q}(\sqrt{2}))^{\times} \cong (\mathbb{Q}(\sqrt{3}))^{\times}$
QUESTION [8 upvotes]: How to show that $$(\mathbb{Q}(\sqrt{2}))^{\times} \cong (\mathbb{Q}(\sqrt{3}))^{\times}$$ where $(\mathbb{Q}(\sqrt{2}))^{\times}$ is multiplicative group of $\mathbb{Q}(\sqrt{2})$.
Mapping $a+b\sqrt{2} \mapsto a+b\sqrt{3}$ is good for showing that additive group of $\mathbb{Q}(\sqrt{2})$ is isomorphic with additive group of $\mathbb{Q}(\sqrt{3})$. And  I showed that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are not isomorphic. But for multiplicative part I couldn't find good mapping.

REPLY [4 votes]: Hints:

${\bf Q}^\times\cong \{\pm1\}\times\bigoplus_p{\bf Z}$, where the $p$th coordinate is the power of $p$ in the prime factorization of a rational into positive/negative prime powers, and the sign is the rational's sign.
Both ${\bf Q}(\sqrt{2})$ and ${\bf Q}(\sqrt{3})$ have unique factorization (or more precisely, their rings of integers)