If $A$ is a Noetherian ring, $M$ is a finitely generated module,
$I$ is an ideal of $A$, and $\hat{A}$ is the $I-adic$ completion of $A$,
then we know that $\hat{A}\otimes\_{A}M\cong\hat{M}$.


Also in Atiyah&Macdonald, there is a remark on Page 109 that the functor $M \mapsto \hat{M}$ is not exact without assuming $M$ finitely generated. but the functor $M \mapsto \hat{A}\otimes\_{A}M$ is always exact. 


How to prove this assertion?


And what is an example of the breakdown of exactness of $M \mapsto \hat{M}$ when $M$ is not finitely generated?


(This is not a homework problem)



On exactness of the functors $M \mapsto \hat{M}$ and $M \mapsto \hat{A}\otimes_{A}M$
I know that this was answered at the comments, but I want to emphasize that actually, the completion functor is not exact from either side, not from the left and not from the right. The only thing that is true (over noetherian or non-noetherian rings) is that completion preserves surjection:


Let $k$ be a field, $A=k[[t]]$, $I=(t)$. In <http://arxiv.org/pdf/0902.4378v4.pdf> example 3.20, there is an example of a short exact sequence of $A$-modules 
$0 \to P \to Q \to M \to 0$, such that after completion, the resulting sequence
$0 \to \hat{P} \to \hat{Q} \to \hat{M} \to 0$ is not even exact at $\hat{Q}$. Thus, completion is no even exact at the middle!


As for tensoring with the completion ring, this is of course a question of flatness. It is know clasically that if $A$ is noetherian then the completion map is flat.


More generally, the completion map does not have to be flat. If $A$ is a ring which is not coherent, then taking $B=A[x]$ and $I=(x)$ will give you an example of a ring with non-flat completion. Note however that in this case, not only your ring is non-noetherian, but also its completion is non-noetherian. It is an open question to me if there is a ring with noetherian completion such that the completion map is not flat.


