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

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)

• What is $\hat A$? Commented Nov 7, 2013 at 2:49
• for a certain ideal $I$, the $I-adic$ completion of $A$
– Alex
Commented Nov 7, 2013 at 3:01
• What an ugly rendering of $\hat A$ on my computer, never seen anything like it. My pdf's don't render $\hat$ that way. Commented Nov 7, 2013 at 3:08
• @PatrickDaSilva, that's because you probably use the HTML-CSS method to render MathJax. You can switch rendering to SVG, then your hats will be fine. I'm not sure why the HTML-CSS method renders hats so badly though. Commented Nov 7, 2013 at 16:34

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

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

• shouldn't the completion of $\mathbb{Z}$ at $p$ be the $p-adic$ integers, which are $\Sigma_{0}^{\infty}a_n p^{n}$?
– Alex
Commented Nov 7, 2013 at 3:22
• what do you denote by $\mathbf{Z}_p$?
– Alex
Commented Nov 7, 2013 at 3:23
• Dear @Alex $\mathbf Z_p$, or $\mathbb Z_p$ is standard notation for the $p$-adic integers! :) Commented Nov 7, 2013 at 3:31
• oh, sorry. I thought you meant localization.
– Alex
Commented Nov 7, 2013 at 3:47
• @Alex : In this case, it's the same. $\mathbb Z_p \otimes_{\mathbb Z} \mathbb Z$ or $(\mathbb Z - (p))^{-1} \mathbb Z$ are the same objects. Commented Nov 7, 2013 at 14:01