How do I prove if the functor is right exact or left exact? Hello I have the following question:

I need to show if the functor $\Bbb{Q}\otimes_R-$ from $\Bbb{Z}$-modules to $\Bbb{Q}$ vector spaces are left exact, right exact or even nothing.

I somehow struggle understanding this functor so I mean before I can figure out if we have a type of exactness I should understand what this functor does. Could maybe someone explain me this and give me a hint if we have some tipe of exactness so that I know where I need to look in more detail. You don't need to tell me why we have this tipe but only a hint so that I'm not totally lost and I get a feeling.
Thanks for your help
 A: Let $R$ be a ring and $M$ be an $R$-module. You know that the tensor product with $M$ has a universal property. Use this to show that any $R$-linear map $f:N\rightarrow N'$ induces an $R$-linear map $M\otimes_R f: M\otimes_R N \rightarrow M\otimes_R N'$. If you want to have a concrete assignment: it is the map given by $m\otimes n \mapsto m \otimes f(n)$. It is then easy to check that this makes $M\otimes_R -$ into a functor from $\mathsf{Mod}_R$ to $\mathsf{Mod}_R$.
If instead of an $R$-module you use an $R$-algebra $A$, something interesting happens: $R$-modules of the form $A\otimes_R N$ have a canonical $A$-module structure with multiplication given by $a\cdot (a‘\otimes n) = (aa‘)\otimes n$. Hence we may regard $A\otimes_R -$ as a functor from $\mathsf{Mod}_R$ to $\mathsf{Mod}_A$.
Now we recall some properties of the tensor product:

*

*By the usual discussion we have a canonical isomorphism of $R$-modules
$$\begin{array}{rcl}
A\otimes_R (M\oplus N) & \overset{\cong}{\longrightarrow} & (A\otimes_R M)\oplus(A\otimes_R N)\\
a \otimes (m,n) & \longmapsto & (a\otimes m, a\otimes n)
\end{array}$$
which is in fact an isomorphism of $A$-modules (with the canonical $A$-module structure described above). This shows that the functor $A\otimes_R -$ is additive.


*By the explicit description of the morphism $M\otimes f$ given above it is easy to see that if $f$ is surjective, so is $M\otimes f$. Together with the additivity this makes $A\otimes_R-$ right exact.
Note that up until now we could do everything in full generality. But for left exactness we need more assumptions. In general neither of the functors $M\otimes_R -$ or $A\otimes_R -$ is left exact. A counterexample is $R=\Bbb Z$, $f:\Bbb Z \rightarrow \Bbb Z, x\mapsto 2x$, $A=\Bbb Z/2\Bbb Z$, since $\Bbb Z/2\Bbb Z\otimes_{\Bbb Z} \Bbb Z = \Bbb Z$ and $\Bbb Z/2\Bbb Z \otimes_\Bbb{Z}=0$.
One says that a module $M$ respectively an algebra $A$ is flat over $R$, of the corresponding tensor functor $M\otimes_R -$ respectively $A\otimes_R -$ is left exact. So in your example we just need to figure out, whether $\Bbb Q$ is flat over $\Bbb Z$. This is true since $\Bbb Q$ is a localization of $\Bbb Z$ and localization is exact.
