Its known that "inverse limit is not exact". Matsumura in his book Commutative Ring Theory, page 272, gives an example for this. I can not understand how he proves that inverse limit of $Z$ is zero. Can you please help me?
Is there another (more simple) example that shows that "inverse limit is not exact"?

thank you


1 Answer 1


For completeness, the example mentioned is the diagram $$\require{AMScd} \begin{CD} @. \vdots @. \vdots @.\vdots\\ @. @VV{p}V @VV{p}V @VV{p}V\\ 0 @>>> \mathbf{Z} @>{n}>> \mathbf{Z} @>>> \mathbf{Z}/(n) @>>> 0\\ @. @VV{p}V @VV{p}V @VV{p}V\\ 0 @>>> \mathbf{Z} @>{n}>> \mathbf{Z} @>>> \mathbf{Z}/(n) @>>> 0\\ @. @VV{p}V @VV{p}V @VV{p}V\\ 0 @>>> \mathbf{Z} @>{n}>> \mathbf{Z} @>>> \mathbf{Z}/(n) @>>> 0 \end{CD}$$ where $n$ and $p$ are coprime.

I think what might be confusing is that you have to take an inverse limit of the sequence of maps $\mathbf{Z} \overset{p}{\leftarrow} \mathbf{Z} \overset{p}{\leftarrow} \cdots$, which is different from the inverse limit of $\mathbf{Z} \overset{\mathrm{id}}{\leftarrow} \mathbf{Z} \overset{\mathrm{id}}{\leftarrow} \cdots$. In particular, you are right that the inverse limit of the latter will be $\mathbf{Z}$.

On the other hand, $$\varprojlim \left(\mathbf{Z} \overset{p}{\leftarrow} \mathbf{Z} \overset{p}{\leftarrow} \cdots\right) = \left\{(a_0,a_1,\ldots,a_n,\ldots) \in \prod_{i=0}^\infty \mathbf{Z}\ \middle\vert\ p^{j-i}a_j = a_i\ \text{for all}\ j \ge i \right\} = 0$$ since if, say, $a_i \ne 0$, then it must be divisible by every power of $p$. If you haven't seen the concrete description of the inverse limit, see for example Atiyah-Macdonald, p. 103. Thus, the inverse limit of the diagram above is $$0 \longrightarrow 0 \longrightarrow 0 \longrightarrow \mathbf{Z}/(n) \longrightarrow 0$$ since the right vertical arrows are all isomorphisms; this sequence is obviously not exact.

Another example is the following from Atiyah-Macdonald, Exc. 10.2: $$\require{AMScd} \begin{CD} @. \vdots @. \vdots @.\vdots\\ @. @VV{p}V @| @VVV\\ 0 @>>> \mathbf{Z} @>{p^n}>> \mathbf{Z} @>>> \mathbf{Z}/p^n\mathbf{Z} @>>> 0\\ @. @VV{p}V @| @VVV\\ @. \vdots @. \vdots @.\vdots\\ @. @VV{p}V @| @VVV\\ 0 @>>> \mathbf{Z} @>{p^2}>> \mathbf{Z} @>>> \mathbf{Z}/p^2\mathbf{Z} @>>> 0\\ @. @VV{p}V @| @VVV\\ 0 @>>> \mathbf{Z} @>{p}>> \mathbf{Z} @>>> \mathbf{Z}/p\mathbf{Z} @>>> 0 \end{CD}$$ The inverse limit is $$0 \longrightarrow 0 \longrightarrow \mathbf{Z} \longrightarrow \hat{\mathbf{Z}}_p \longrightarrow 0$$ where the left and middle columns are as above, and the right column is the example right above yours in Matsumura, p. 272; this sequence is not exact.

  • $\begingroup$ :can I ask you what are you reading for homological algebra? $\endgroup$
    – pink floyd
    Feb 19, 2015 at 18:28
  • $\begingroup$ I'm personally reading Weibel's An introduction to homological algebra. the discussion in §3.5 seems relevant, and in particular, Example 3.5.5 is my second example. $\endgroup$ Feb 19, 2015 at 23:14
  • $\begingroup$ @ takumi murayama:thank you $\endgroup$
    – pink floyd
    Feb 20, 2015 at 7:24

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