I am learning local cohomology from Hartshorne’s Local Cohomology book. My question is about the notion of essentially zero inverse system of abelian groups, which is defined to be an inverse system of abelian groups $(M_{m})$ indexed by the non-negative integers, such that for every $m$, there is some integer $m’\geq m$, such that $M_{m’}\rightarrow M_{m}$ is the zero map.

Hartshorne’s book has the following remark, which I have trouble verifying:

If we have an exact sequence of inverse systems

$$0\rightarrow (M’_{m})\rightarrow (M_{m})\rightarrow (M''_{m})\rightarrow 0,$$

then the middle one is essentially zero if and only if the two outside ones are essentially zero.

I could only show that if the middle inverse system is essentially zero, then so are the two outside inverse systems.

The idea is to use a commutative diagram of the two short exact sequences induced by the zero map $M_{m’}\rightarrow M_{m}$.

So my question is, how to show that if the two outside inverse systems are essentially zero, then so is the middle one? I have tried to apply the same idea as above, but did not get a proof. Thank you so much for your kind help.

  • $\begingroup$ TeX note: be sure to enter a double prime as two apostrophes (or ^{\prime\prime}, if for some reason you like that better). Compare $M”_m$ M”_m to $M''_m$ M''_m. I edited accordingly. $\endgroup$
    – LSpice
    Commented Feb 6, 2023 at 15:11
  • $\begingroup$ Thank you so much for your kind help. $\endgroup$
    – Boris
    Commented Feb 6, 2023 at 15:13
  • $\begingroup$ I deleted my original comment, where I misunderstood that the outer modules themselves were eventually $0$. This question is more interesting (though I suspect still a formal check). $\endgroup$
    – LSpice
    Commented Feb 6, 2023 at 15:31
  • $\begingroup$ Thank you very much for your comments. Hope I will be able to solve this problem with some help. $\endgroup$
    – Boris
    Commented Feb 6, 2023 at 15:58
  • $\begingroup$ Too many primes... Why not $0\to X_n\to Y_n\to Z_n\to 0$? Given $n\in \mathbb N$, first choose $m>n$ such that $X_m\to X_n$ is $0$ and then $k>m$ such that $Z_k\to Z_m$ is $0$. Then $Y_k\to Y_n$ is $0$ by a very simple diagram chase. $\endgroup$ Commented Feb 6, 2023 at 17:49

2 Answers 2


I don't like indexing with primes, so let's consider an exact sequence of inverse systems:

$$ 0 \to (A_m) \to (B_m) \to (C_m) \to 0 $$

Now we are assuming that $(A_m)$ and $(C_m)$ are essentially zero inverse systems, which means that for a given $m$ we can find a larger $m'$ so that both $A_{m'} \to A_m$ and $C_{m'} \to C_m$ are the zero map.

At this point we get a map of short exact sequences:

$\require{AMScd}$ \begin{CD} 0 @>>> A_{m'} @>i_{m'}>> B_{m'} @>>> C_{m'} @>>> 0 \\ @. @VV 0 V @VVf_{m', m} V @VV 0 V @. \\ 0 @>>> A_{m} @>>i_m> B_{m} @>>> C_{m} @>>> 0 \\\end{CD}

Now the tricky point is that it does not follow that $f_{m', m}$ is the zero map. There can certainly be maps of short exact sequences where the outside maps are zero, but the central map is non-zero. However we can record two facts.

First, by the commutivity of the diagram and the fact that the rows are exact we see that the image $f_{m', m}(B_{m'})$ is contained in the image $i_m(A_m)$. Second, by the commutativity of the diagram, the image of $i_{m'}(A_{m'})$ lies in the kernel of $f_{m', m}$.

Summarizing, the sequence $(B_m)$ satisfies the property that for each $m$ there exists an $m' \geq m$ such that

  1. $f_{m', m}(B_{m'}) \subseteq i_m(A_m)$
  2. $i_{m'}(A_{m'}) \subseteq \ker f_{m', m}$

Now given $m$ we apply this property twice, first to $m$ to get $m'$ and then to $m'$ to get $m''$.

Then the image of $B_{m''}$ under $f_{m'', m'}$ lies in the image of $A_{m'}$, which in turn lies in the kernel of $f_{m', m}$. Thus the map

$$f_{m'', m} = f_{m', m} \circ f_{m'', m'}: B_{m''} \to B_m$$

is zero.

  • $\begingroup$ Thank you so much for your kind help and detailed explanations. They are very clear and insightful. $\endgroup$
    – Boris
    Commented Feb 6, 2023 at 20:29

This is the proof mentioned in my comment which was considered doubtful: Let the exact sequence be $0\to X\stackrel{i}{\to} Y\stackrel{q}{\to}Z\to 0$ and denote the linking maps of the inverse systems $\alpha_m^n:X_m\to X_n$, $\beta_m^n:Y_m\to Y_n$ and $\gamma_m^n:Z_m\to Z_n$ for $n\le m$.

Given $n\in\mathbb N$, choose $m>n$ such that $\alpha_m^n=0$ and then $k>m$ such that $\gamma_k^m=0$. Then $\beta_k^n=0$. Indeed, for $y\in Y_k$, we have $q_m(\beta_k^m(y))=\gamma_k^m(q_k(y))=0$ so that there is $x\in X_m$ with $\beta_k^m(y)=i_m(x)$. This implies $$\beta_k^n(y)=\beta_m^n(\beta_k^m(y))=\beta_m^n(i_m(x))=i_n(\alpha_m^n(x))=i_n(0)=0.$$

We only used the exactness at $Y$.

  • $\begingroup$ Thank you so much for your kind help and explanations. Now I understand your proof by diagram chase. $\endgroup$
    – Boris
    Commented Feb 7, 2023 at 0:37

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