Recently, some convergence problems I have been considering led me to look at additive maps of the torus (which for me is the additive group $T := \mathbb{R}/\mathbb{Z}$).
A map $f:\ T \to T$ is additive if $f(\alpha + \beta) = f(\alpha) + f(\beta)$. Other names that seem to make sense would be $\mathbb{Z}$-linear map or group homomorphism, depending on how one wants to view things. The simplest examples are of course the linear maps $f(\alpha) = c \alpha$, $c \in \mathbb{Z}$.
If we were talking about additive maps $f:\ \mathbb{R} \to \mathbb{R}$, then the situation is fairly well understood. One can show that there exists a basis of $\mathbb{R}$ as a $\mathbb{Q}$-linear space. Let $\{\alpha\}_{i \in I}$ be such a basis, and choose arbitrary reals $\{\beta\}_{i \in I}$ : then there is a unique additive map $f$ such that $f(\alpha_i ) = \beta_i$ (more explicitly, $f(\sum_{i \in I} q_i \alpha_i) = \sum_{i \in I} q_i \beta_i$, where $q_i \in \mathbb{Q}$ and $q_i = 0$ except finitely many $i$). Moreover, any additive map arises in this way, so we have a complete characterisation (modulo the fact that a $\mathbb{Q}$-linear basis of $\mathbb{R}$ is not something easy to come by).
If we consider an additive map $f:\ \mathbb{R} \to \mathbb{R}$ such that $f(1) \in \mathbb{Z}$, then $f$ factors through the quotient with $\mathbb{Z}$, giving rise to the additive map $[f] :\ T \to T$ given by $[f]([\alpha]) = [f(\alpha)]$. Hence, there exists a fair number of additive maps of the torus.
Question: Does every additive map $T \to T$ arise in the way just described? Is there a general description of additive maps $T \to T$, like the one for $\mathbb{R}$? [Affirmative answer to the second question probably resolves both].
Note: if $f:\ T \to T$ is additive, and $\{\alpha\}_{i \in I}$ is a basis of $\mathbb{R}$ over $\mathbb{Q}$ as above, then clearly $f(\sum_i a_i \alpha_i) = \sum_i a_i f(\alpha_i)$ as long as $a_i$ are integers. However, it is not the case that one can select $\alpha_i$ such that each $\beta \in T$ is a unique integral combination of $\alpha_i$ (in other words, $T$ is not a free group). It is true that $f(\sum_i q_i \alpha_i) = \sum_i q_i f(\alpha_i) + Q$ with $q_i \in \mathbb{Q}$ and some $Q \in \mathbb{Q}$ dependent on $q_i$ (since we have the additional rational term anyway, we may understand rational multiplication $q_i f(\alpha_i)$ in any way we like.).