How can $\mathbb{R}/\mathbb{Z}$ be a free $\mathbb{R}$-module?

Question

I'm a little confused here.

On the one hand, $\mathbb{R}/\mathbb{Z}$ must be free as an $\mathbb{R}$-module, since any module over a field is free (being a vector space it has a basis).

One the other hand, it can't be free, as it has torsion - e.g. $2(0.5 + \mathbb{Z})=0$.

What's wrong with my thinking here?

Answer 1

One big problem here is that you are trying to form the quotient module $\Bbb R/\Bbb Z$, even though $\Bbb Z$ is not an $\Bbb R$-submodule of $\Bbb R$. So this quotient doesn't make sense as an $\Bbb R$-module (although it does make sense as an abelian group).

Answer 2

Because $\mathbb{R}/\mathbb{Z}$ isn't an $\mathbb{R}$-module in the first place.

(It is a $\mathbb{Z}$-module - that is, an abelian group - but not an $\mathbb{R}$-module.)

The issue is that when we try to pass from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Z}$, we break scalar multiplication: letting $q:\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$ be the "quotient map" of sets, note that $q(1)=q(0)$ but $q({1\over 2}\cdot 1)\not=q({1\over 2}\cdot 0)$, so multiplication by the scalar ${1\over 2}$ isn't well-defined in $\mathbb{R}/\mathbb{Z}$.

Indeed, only multiplication by integer scalars is unproblematic - writing "$[x]_\mathbb{Z}$" for $\{x-z:z\in\mathbb{Z}\}$, we get:

• If $z\in\mathbb{Z}$ and $[x]_\mathbb{Z}=[y]_\mathbb{Z}$ then $zx-zy=z(x-y)\in\mathbb{Z}$ so $[zx]_\mathbb{Z}=[zy]_\mathbb{Z}$.

• If $r\in\mathbb{R}\setminus\mathbb{Z}$, then even though $[0]_\mathbb{Z}=[1]_\mathbb{Z}$ we'll unfortunately have $[r\cdot 0]_\mathbb{Z}\not=[r\cdot 1]_\mathbb{Z}$ (since $r\cdot0\in\mathbb{Z}$ but $r\cdot 1\not\in\mathbb{Z}$).