When is a field "compatible" with an abelian group?

Question

Let $F$ be a field and $G$ an abelian group. We say that $F$ and $G$ are "compatible" if and only if there exists a function $M:F\times G\to G$ such that

1. $M(\lambda,gh)=M(\lambda,g)M(\lambda,h)$ for all $\lambda\in F$, $g,h\in G$

2. $M(\lambda+\mu,g)=M(\lambda,g)M(\mu,g)$ for all $\lambda,\mu\in F$, $g\in G$

3. $M(\lambda\mu,g)=M(\lambda,M(\mu,g))$ for all $\lambda,\mu\in F$, $g\in G$

4. $M(1,g)=g$ for all $g\in G$

5. $M(0,g)=\varepsilon$ for all $g\in G$, where $\varepsilon$ is the identity of $G$

If $F$ and $G$ are compatible, then $(F,G)$ (or $\{F,G,M\}$, or $\langle F,+_F,\cdot_F,G,+_g,M\rangle$, or however else you care to denote it) forms a vector space, whose field is $F$, whose vectors are $G$ with vector addition given by the group operation, and scalar multiplication given by $M$.

In fact, this is the definition of a vector space - a vector space consist precisely of a field and an abelian group related by a function like the one described.

Question: What methods can be used to efficiently decide the compatibility of a given field with an arbitrary abelian group?

There is a trivial case. If we take $F^X$ to be the group of functions from $X$ to $F$ with pointwise addition, and $G\cong F^X$, then $F$ and $G$ are compatible by $M(\lambda,\varphi^{-1}(g))(x)=\lambda \varphi^{-1}(g)(x)$, where $\varphi:F^X\to G$ is any group isomorphism. Every vector space for which I have a name is of this form, but, if I'm not mistaken, one can easily construct one which isn't by matching a group with an involution to a field whose additive group has none.

Answer 1

What might be called the fundamental theorem of linear algebra - "Every vector space has a basis" - solves this problem quickly: if $V$ is a vector space over $k$, then $V$ as an abelian group is isomorphic to some direct sum of the additive group of $k$. Put another way:

Suppose we have an abelian group $G$ and a field $k$. Then $G$ can be equipped with the structure of a $k$-vector space iff for some (possibly infinite) cardinal $\mu$ we have $G\cong \bigoplus_{i<\mu}k_+$, where $k_+$ is the additive group of $k$.

Note the direct sum rather than direct product here: the point is that if $B$ is a basis for a vector space $V$, each element of $V$ only "uses" finitely many elements of $B$.

Also, note that things are more interesting for modules (= just require that $k$ be a ring, not necessarily a field): for example, $\mathbb{Q}$ is a $\mathbb{Z}$-module in an obvious way despite not being a direct sum of copies of $\mathbb{Z}$.