Prove that $\mathbb{Z}$ is not isomorphic to additive group of any vector space over any field. Prove that $\mathbb{Z}$ is not isomorphic to additive group of any vector space over any field.
Proof.
Surpose that: $\phi : (A, +) \rightarrow  \mathbb{Z} $ is an isomorphism. Then there is some $r∈A $ such that $ϕ(r)=1_\mathbb{Z}$.
We have
$$ 1_\mathbb{Z}=ϕ(r)=ϕ(2(r/2))=2ϕ(r/2)$$
or equivalently that $ϕ(r/2)=\frac{1}{2}$. But this is not in $ \mathbb{Z}$, so there can be no such morphism.
Is my proof above correct? Thanks
 A: Let $F$ be a field and $V$ a (non-trivial) vector space over $F$.
Hints: $V$ contains a subgroup isomorphic to the additive group of $F$.
If $F$ has characteristic $p>0$, then there is no isomorphism because $\mathbb{Z}$ does not contain an element of order $p$.
If $F$ has characteristic $0$, then it contains a subfield isomorphic to the rationals.
$\mathbb{Q}$ is not isomorphic to $\mathbb{Z}$.
A: In a vector space of dimensión strictly larger than one there exist non-trivial subgroups which intersect trivially. As that is not the case in the integers, if Z is the underlying abelian group of a vector space, the latter is of dimensión 1. Of course, we see that in that case Z is in fact isomorphic to the underlying additive group of a field.
As there is no torsion in Z, the field must have characteristic zero. But the additive group of a field of characteristic zero is divisible and Z isn't. 
A: If the underlying field has characteristic $2$, then $2 = 0$ so you cannot look at $r/2$. You could deal with that case separately, though, by dividing by something else.
