rational vector space as uniquely divisible abelian group

Proposition

Let \(M\) be an abelian group.
TFAE:

  1. \(M\) admits a unique structure as \(\mathbb{Q}\)-vector space
  2. \(M\) admits a structure as \(\mathbb{Q}\)-vector space
  3. for each \(\frac{1}{q} \in \mathbb{Q} \setminus \{0\}\) and each \(m \in M\) there exists a unique \(m'\) such that
\begin{align*} \sum_{i=1}^q m' = m \end{align*}

In that case \(m' = \frac{1}{q} m\)

Proof

1) \(\implies\) 2)

special case

2) \(\implies\) 3)

Existence: follows from \(m' := \frac{q}{1} m\) where \(\sum_{i=1}^q\) (todo)
For uniqueness: let \(m',\tilde{m}'\) be such elements, then

\begin{align*} q \cdot m' = q \cdot \tilde{m}' && \vert \cdot q^{-1} \\ m' =& \tilde{m}' \end{align*}

3) \(\implies\) 1)

for \(\frac{p}{q} \in \mathbb{Q} \setminus \{0\}\) and \(m \in M\) define

\begin{align*} \frac{p}{q} m = \sum_{i=1}^p (m') \\ \end{align*}

where \(m'\) is the unique element satisfying \(\sum_{i=1}^q m' = m\)
then one can check, that this defines a vectorspace structure on \(M\)

alternative

Date: nil

Author: Anton Zakrewski

Created: 2026-01-12 Mo 17:22