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In this question, it is asked whether the additive group of a ring can be isomorphic to its group of units. The answer is yes: the trivial ring is the easiest example. However, the top-voted answer asserts that $\mathbb R\times \mathbb Z/2\mathbb Z$ (under addition) is isomorphic to $(\mathbb R\times \mathbb Z/2\mathbb Z)^\times$. Why is this the case?

I can verify that the units of the ring $\mathbb R\times \mathbb Z/2\mathbb Z$ are of the form $(x,1)$, where $x\neq0$, so that there is an isomorphism between $(\mathbb R\times\mathbb Z/2\mathbb Z)^\times$ and $\mathbb R^\times$ which takes $(x,1)$ to $x$, but I cannot find an isomorphism $\mathbb R^\times\to\mathbb R\times \mathbb Z/2\mathbb Z$.

Joe
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  • "There is an isomorphism between $\Bbb R\times\Bbb Z_2$ and $\Bbb R^\times$ which takes $(x,1)$ to $x$." This is a false statement. You just described an isomorphism between $(\Bbb R\times\Bbb Z_2)^\times$ and $\Bbb R^\times$, not an isomorphism between $(\Bbb R\times\Bbb Z_2,+)$ and $(\Bbb R^\times,\cdot)$. An isomorphism is in particular a function, not a "partial" function, so you described a function which does not have all of $\Bbb R\times\Bbb Z_2$ as its domain. – anon Apr 12 '23 at 22:07
  • @runway44: Thanks, there was a typo in my post. I meant an isomorphism between $(\mathbb R\times \mathbb Z/2\mathbb Z)^\times$ and $\mathbb R^\times$. – Joe Apr 12 '23 at 22:08

1 Answers1

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The isomorphism $\mathbb{R}\times\mathbb{Z}_2\to\mathbb{R}^\times$ is given by $(x,s)\mapsto(-1)^s e^x$.

The "parity bits" of $\mathbb{Z}_2$ corresponds to two connnected components of $\mathbb{R}^\times$ (the positive or negative numbers), and exponentials famously turn addition into multiplication, i.e. $e^{x+y}=e^x\cdot e^y$.

The inverse isomorphism $\mathbb{R}^\times\to\mathbb{R}\times\mathbb{Z}_2$ is then $r\mapsto(\ln|r|,\,\mathrm{sgn}\,r)$.

anon
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