
  Let
  $A\stackrel{\alpha}{\rightarrow}B\stackrel{\beta}{\rightarrow}C\rightarrow 0$ a exact sequence of left $R$-modules and $M$ a left $R$-module ($R$
  any ring).
  
  I am trying to prove that the induced sequence $$A\otimes_R M\xrightarrow{\alpha\otimes Id}B\otimes_R M\xrightarrow{\beta\otimes Id}C\otimes_R M\rightarrow 0$$ is
  exact.


The part I have trouble with is that $\ker{\beta\otimes Id}\subset\text{im }{\alpha\otimes Id}$. 

If we had $$\beta(b)\otimes m=0 \text{ if and only if } \beta(b)=0\text{ or }m=0,$$
we could easily conclude using the exactness of the original sequence. However, it is false, right ? (I think of $C_3\otimes \mathbb{Z}/2\mathbb{Z}$, we have $g^2\otimes 1=g\otimes 2=g\otimes 0=0$, where $g$ is a generator of $C_3$.)

I can't see how to proceed then... When a tensor $c\otimes m$ is zero, what can we say on $c$ and $m$ in general ?

The proof mentioned by Frederik and Loronegro is great because it provides a first example of how it can be useful to know that two functors are adjoint: left adjoints are right exact. However, you can also argue as follows. Let $D$ be the image of $\alpha \otimes \operatorname{id}$. You get an induced map $(B \otimes M)/D \to C \otimes M$. Let's try to define an inverse: if $(c, m) \in C \times M$ then choose a $b \in B$ such that $\beta(b) = c$, and send $(c, m)$ to $b \otimes m \bmod D$. You can check that this is well defined using the exactness of the original sequence.