Is $\mathbb Z$ a $\mathbb Q$- module? Could we define proper $\mathbb Q$ module structure on ring of integer $\mathbb Z$?
Any comments are welcome.
 A: If it would be (call the module multiplication '$\cdot$') then what should $1/2 \cdot 1 = a$ be? It would be an $a \in \mathbb{Z}$ such that $a+a = 2\cdot a = 2 \cdot (1/2 \cdot 1) = 1 \cdot 1 = 1$, pretty absurd, right?
In general, a $\mathbb{Q}$-module $M$ is a vector space over $\mathbb{Q}$ so we have a lot of structure. For example, every $q \in \mathbb{Q}$ with $q\not=0$ gives rise to an automorphism $M \rightarrow M$ which sends $m$ to $q\cdot m$ (why is this an automorphism). This in particular implies that $m \mapsto 2\cdot m = m+m$ and $m\mapsto 3\cdot m$ etc. are all automorphisms of $M$, which is clearly not the case for $\mathbb{Z}$. 
A: Not while respecting the usual addition of $ℤ$.
Either you give up all natural structure of $ℤ$, treat it merely as any countable set $A$ and use some bijection $ρ \colon A → ℚ$ to artificially copy all the structure from $ℚ$ to $A$ – which seems rather pointless – …
… or you try to find a $ℚ$-module structure on $(ℤ,+)$. But then you’d fail: Say you had a scalar multiplication from $ℚ × ℤ → ℤ$. From the definition of modules, we have in particular the following two properties.


*

*$∀x ∈ ℤ$: $1·x = x$.

*$∀r,s ∈ ℚ, ∀x ∈ ℤ$: $(r + s)·x = r·x + s·x$.


Now, what happens if you set $r = s = \frac 1 2$ in $ℚ$ and $x = 1$ in $ℤ$?
A: Actually a lot stronger result holds: There is no $\mathbb Q$-module structure on any finitely generated $\mathbb Z$-algebra, i.e. on any finitely generated ring.
