on exactness of the functors $M \mapsto \hat{M}$ and $M \mapsto \hat{A}\otimes_{A}M$

Question

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

Also on 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)

Answer 1

One example is the sequence of abelian groups

$$0 \to \mathbf Z \to \mathbf Q \to \mathbf Q/\mathbf Z \to 0.$$

(Remark that neither $\mathbf Q$ nor $\mathbf Q/\mathbf Z$ is finitely-generated as a $\mathbf Z$-module.)

If we complete this at $p$, we get the sequence

$$0 \to \mathbf Z_p \to 0 \to 0 \to 0$$

which is obviously not exact. However, if we had tensored with the p adic integers $\mathbf Z_p$ instead, we would have gotten the exact sequence

$$0 \to \mathbf Z_p \to \mathbf Q_p \to \mathbf Q_p/\mathbf Z_p \to 0.$$