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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)

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    $\begingroup$ For the counterexample, $A =\mathbb Z$, $I=p$, $M= \mathbb Q$. $\endgroup$
    – Will Sawin
    Commented Nov 7, 2013 at 4:23
  • $\begingroup$ I don't understand the counterexample. A counterexample should consist of an exact sequence, not just a module. $\endgroup$
    – Steven Landsburg
    Commented Nov 7, 2013 at 14:18
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    $\begingroup$ In Will Sawin's setup, the injection $\mathbb{Z}\hookrightarrow \mathbb{Q}$ of $\mathbb{Z}$-modules does not remain injective after taking $p$-adic completions. $\endgroup$
    – Kevin Ventullo
    Commented Nov 7, 2013 at 18:16
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    $\begingroup$ Also, if you look closely on page 109 of A-M, there is actually a proof of the flatness of $\hat{A}$ as an $A$-algebra. $\endgroup$
    – Kevin Ventullo
    Commented Nov 7, 2013 at 18:23
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    $\begingroup$ You don't even have to look closely on page 109 of AM: the remark immediately follows the proposition that $\hat A$ is a flat $A$-algebra. Furthermore, the very first exercise in the section is a counterexample to exactness for non-finitely generated modules. $\endgroup$
    – Jack Huizenga
    Commented Nov 7, 2013 at 18:48

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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.

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