# $I$-adic completion of an exact sequence of finite $A$-modules is again exact

I'm reading Hideyuki Matsumuras Commutative Ring Theory, and I'm having a trouble understanding a statement he makes in theorem 8.7.

For some context, the following two theorems have previously been proven:

1) $A$ noetherian, $M$ finite $A$-module, $N\subset M$ submodule and $I$ an ideal of $A$. Then the $I$-adic topology of $N$ coincides with the topology induced by the $I$-adic topology of $M$ on the subspace $N$.

And

2) In the $I$-adic topology the sequence $0\rightarrow \hat{N} \rightarrow \hat{M} \rightarrow \hat{M/N}\rightarrow 0$ is exact.

Now I'm reading, that from these two theorems, we can conclude that

Let $A$ be a noetherian ring and $I$ an ideal. Then the $I$-adic completion of an exact sequence of finite $A$-modules is again exact.

But I don't get why that is the case? I don't understand how we can generalize the fact that if $0\rightarrow {N} \rightarrow {M} \rightarrow {M/N}\rightarrow 0$ is an exact sequence, then taking I-adic completion we get that $0\rightarrow \hat{N} \rightarrow \hat{M} \rightarrow \hat{M/N}\rightarrow 0$ beign exact, to given an arbitrary sequence $0\rightarrow {A} \rightarrow {B} \rightarrow {C}\rightarrow 0$ which is exact, then taking I-adic completion Them $0\rightarrow \hat{A} \rightarrow \hat{B} \rightarrow \hat{C}\rightarrow 0$ is exact?

Now I know that the $I$-adic topology on $M$ can be defined by $\{I^nM\}_{n\in \mathbb{N}}$, the one on $M/N$ by $\{M/(N+I^nM)\}_{n\in \mathbb{N}}$ and the one on $N$ by $\{N/(N\cap I^nM)\}_{n\in \mathbb{N}}$, so I was thinking we somehow could use that to generalize, but I haven't come Any futher.

• Not sure it helps, but if $F$ is an additive functor from finite $A$-modules to abelian groups, then the conditions $$0\to N\to M\to P\to0\text{ exact implies }0\to FN\to FM\to FP\to0\text{ exact}$$ $$N\to M\to P\text{ exact implies }FN\to FM\to FP\text{ exact}$$ are equivalent. – Pierre-Yves Gaillard Apr 23 '18 at 17:15
• The second condition is clearly equivalent to: $F$ preserves the exactness of any exact sequence of finite $A$-modules. – Pierre-Yves Gaillard Apr 23 '18 at 17:30
• Generally, we know that if $A$ is Noetherian and $\hat{A}$ its completion, then for finitely generated $M$, we have$\hat{M} \cong \hat{A} \otimes_A M$ and that $\hat{A}$ is flat over $A$. I saw this in Atiyah and Macdonald in more details I believe, but can't recall the proof right now. – Sheel Stueber Apr 23 '18 at 17:48

If $F$ is an additive functor from finite $A$-modules to abelian groups, then the conditions that

$\bullet\ F$ preserves the exactness of exact sequence of finite $A$-modules of the form $$0\to N\to M\to P\to0,$$ $\bullet\ F$ preserves the exactness of exact sequence of finite $A$-modules of the form $$N\to M\to P,$$ $\bullet\ F$ preserves the exactness of any exact sequence of finite $A$-modules,

are equivalent.

See Exercise 8.17 in this text.

EDIT 1. Here is a version which tries to take into account the OP's comment.

Let $Q$ be an $A$-module, and, for any $A$-module $X$ set $$FX:=Q\otimes_AX.$$ Note that any morphism $$f:X\to Y$$ yields an obvious morphism $$Ff:FX\to FY.$$ In particular, applying $F$ to an exact sequence of finite $A$-modules gives rise to a complex of $A$-modules (that is, a sequence of morphisms such that the composition of any two successive morphisms is $0$). (This complex will not be exact in general.)

Then the conditions that

$\bullet\ F$ preserves the exactness of exact sequence of finite $A$-modules of the form $$0\to N\to M\to P\to0,$$ $\bullet\ F$ preserves the exactness of exact sequence of finite $A$-modules of the form $$N\to M\to P,$$ $\bullet\ F$ preserves the exactness of any exact sequence of finite $A$-modules,

are equivalent.

The case you are interested in is $Q=\widehat A$.

EDIT 2. In fact, if $F$ is as in Edit 1, and if $F$ preserves the exactness of $0\to J\to A$ for all finitely generated ideals $J$, then $F$ preserves the exactness of arbitrary exact sequences of arbitrary $A$-modules.

EDIT 3. I think there is a nano-gap in Matsumura's proof of Theorem 8.7:

Write $FX$ for $\widehat X$. Let $X,Y,Z,U,V,W$ be finite $A$-modules. It is proved in Theorem 8.1 that $$0\to X\to Y\to Z\to0$$ exact implies $$0\to FX\to FY\to FZ\to0$$ exact. Let's denote this property by $(\star)$. But the property used in Theorem 8.7 is that $$X\to Y\to Z\to0$$ exact implies $$FX\to FY\to FZ\to0$$ exact.

It's better (I think) to prove that $(\star)$ implies that $F$ preserves the exactness of any exact sequence of finite $A$-modules.

Assuming $(\star)$, let $U\to V\to W$ be exact. It suffices to show that $FU\to FV\to FW$ is exact.

Setting $$U':=\text{Ker}(U\to V),\ V':=\text{Im}(U\to V)=\text{Ker}(V\to W),\ W':=\text{Im}(V\to W),$$ we get the exact sequences $$\begin{matrix} FU&\to&FV'&\to&0\\ \\ 0&\to&FV'&\to&FV&\to&FW'&\to&0\\ \\ &&&&0&\to&FW'&\to&FW. \end{matrix}$$ It follows that $FU\to FV\to FW$ is exact, as required.

• I'm sorry, but I don't really have any experience in what functors is, and how they work, but does this translate to, that if taking inverse limits of one exact sequence preserves the exact sequence, then taking inverse limit of any exact sequence preserves the exact sequence? – jta Apr 23 '18 at 19:10
• Please see the edit. Don't hesitate to let me know if this is not clear. @Jta – Pierre-Yves Gaillard Apr 23 '18 at 19:56
• So since we've seen that $I$-adic completion preserves the exactness of a concrete (in this case $N/I^nN\rightarrow M/I^nM \rightarrow (M/N)/I^n(M/N)$) short exact sequence of finite $A$-modules, then $I$-adic completion becomes a exact functor, and hence $I$-adic completion preserves the exactness of any short exact sequence of finite $A$-modules? I'm confused since Matsumura shows that $I$-adic completion preserves exactness for a concrete chain, and not an arbitrary one – jta Apr 24 '18 at 15:19
• @Jta - Any exact sequence of the form $$0\to X\to M\to Z\to 0$$ is isomorphic (in an obvious sense) to some exact sequence of the form $$0\to N\to M\to M/N\to 0$$ with $N$ submodule of $M$. – Pierre-Yves Gaillard Apr 25 '18 at 12:40