Isomorphism between $\mathbb{Q}/\mathbb{Z}$ and $\mu_{\infty}(\mathbb{C})$, the set of all $n$th roots of unity.

Question

I think that I only need to show that the function $e(x)=e^{2\pi i x}$ for x integers gives back 1 and for a number which is a rational a which is not a natural number we get a complex number but not a real number, am I correct?

@CameronWilliams probably complex numbers a power of which yields $1$ — Evariste, May 18 '21 at 14:24
Sorry for not including it: https://math.stackexchange.com/questions/4143140/a-question-on-roots-of-unity — MathematicalPhysicist, May 18 '21 at 14:24
Ahh so the collection of all roots of unity? $\mu$ is a funny letter to use since that's very often reserved for measures and sometimes a sub- or superscript $\infty$ shows up in that situation. I couldn't figure out how this was being equated to a measure. Hah. — Cameron Williams, May 18 '21 at 14:25

score 5 · Accepted Answer · answered May 18 '21 at 14:23

5

Hint :

Show that the morphism $\varphi : \mathbb{Q} \rightarrow \mu_{\infty}(\mathbb{C})$ defined by $$\varphi \left(\frac{p}{q} \right) = \exp\left(2i\pi \frac{p}{q} \right)$$

is well-defined, surjective, and that its kernel is exactly $\mathbb{Z}$.

answered May 18 '21 at 14:23

TheSilverDoe

29,720

Isomorphism between $\mathbb{Q}/\mathbb{Z}$ and $\mu_{\infty}(\mathbb{C})$, the set of all $n$th roots of unity.

1 Answers1