Torsion subgroup of $\mathbb{C}^\times$

Question

I need to find the torsion subgroup of the multiplicative abelian group $\mathbb{C}^\times$. This is from a homework assignment sheet, and I'm not sure what the notation $\mathbb{C}^\times$ stands for. I'm assuming it's the group of units. Every complex number has a multiplicative inverse, hence $\mathbb{C}^\times=\mathbb{C}$, so I'm not really sure why this notation is necessary, and it makes me think I've got the wrong idea.

The solutions say the torsion subgroup consists of roots of unity. I don't see why this is the case. The torsion subgroup is all the elements of the module that are annihilated by ring, and I don't see how integer multiples of complex numbers ever give 0.

So I've obvious got the wrong end of the stick! Thanks for any help.

Answer 1

Given a ring $R$, the notation $R^{\times}$ is used to denote the multiplicative group of units in $R$. Recall that $r \in R$ is a unit if there is $s \in R$ such that $rs = sr = 1$. If $R$ is a field, then every non-zero element has a multiplicative inverse, so $R^{\times} = R\setminus\{0\}$. As $\mathbb{C}$ is a field, $\mathbb{C}^{\times} = \mathbb{C}\setminus\{0\}$.

An element $g$ of a multiplicative group $G$ is a torsion element if there is $n \in \mathbb{N}\setminus\{0\}$ such that $g^n = 1$. So the torsion elements of $\mathbb{C}^{\times}$ are those complex numbers $z$ such that $z^n = 1$ for some $n$. That is, the torsion subgroup of $\mathbb{C}^{\times}$ is precisely the roots of unity.