This is an exercise from Chapter 3 of Golan's linear algebra book.
Problem: Show $\mathbb{Z}$ is not a vector space over a field.
Solution attempt:
Suppose there is a such a field and proceed by contradiction. I will write multiplication $FV$, where $F$ is in the field and $V$ is an element of $\mathbb{Z}$.
First we rule out the case where the field has characteristic 2. We would have
$$0=(1_F+1_F)1=1_F1+1_F1=2$$
a contradiction.
Now, consider the case where the field does not have characteristic 2. Then there is an element $2^{-1}_F$ in the field, and
$1=2_F(2^{-1}_F1)=2^{-1}_F1+2^{-1}_F1$
Now $2^{-1}_F1\in\mathbb{Z}$ as it is an element of the vector space, but there is no element $a\in\mathbb{Z}$ with $2a=1$, so we have a contradiction.
Is this correct?
Best Answer
The answer is, "Yes."
If $V$ is a vector space over a field of positive characteristic, then as an abelian group, every element of $V$ has finite order. If $V$ is a vector space over a field of characteristic $0$, then as an abelian group, $V$ is divisible. The abelian group $\mathbb Z$ has neither of these properties.