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I am studing the book "Rings and Categories of modules" written by Frank W. Anderson and Kent R. Fuller. On page 124, I am at a loss for an example.

Example. The abelian group $\mathbb{Z}$ is finitely generated but not finitely cogenerated. The abelian group $\mathbb{Z}_{p^\infty}$ is finitely cogenerated but not finitely generated.

I post my effort here.

(1).We can regard them as modules over $\mathbb{Z}$.

(2).$\mathbb{Z}$ is finitely generated is clear.

(3). By the chain $0 \subset \left({1 \over p}\mathbf{Z}\right)/\mathbf{Z} \subset \left({1 \over p^2}\mathbf{Z}\right)/\mathbf{Z} \subset \left({1 \over p^3}\mathbf{Z}\right)/\mathbf{Z} \subset \cdots$ of $\mathbb{Z}_{p^\infty}$. We know $\mathbb{Z}_{p^\infty}$ as $\mathbb{Z}$-module is Artinian but not Noetherian.

Any help will be appreciated.

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Ad (1): You mean you regard them both as modules over ${\mathbb Z}$, no?

Ad (3): The chain you provide indeed shows that ${\mathbb Z}_{p^{\infty}}$ is not finitely generated, but to show that it is artinian and hence finitely cogenerated, you still need to note that you've actually listed all submodules.

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  • $\begingroup$ Thank you for your answer. I regard them both as modules over $\mathbb{Z}$. I know the fact that a module $M$ is artinian iff every factor module of $M$ is finitely cogenerated. I don't know why $\mathbb{Z}_{p^\infty}$ is artinian and hence finitely cogenerated? $\endgroup$ – Daisy Apr 3 '17 at 7:45
  • $\begingroup$ @Daisy: Alternatively, you can say that a module is finitely cogenerated if every descending chain of submodules gets stationary. The point is now that the list of submodules that you have provided for ${\mathbb Z}_{p^{\infty}}$ already lists all submodules - and that it's obvious that there's no infinite decreasing chain in it. Does that help? $\endgroup$ – Hanno Apr 4 '17 at 5:49
  • $\begingroup$ @Daisy Does that help? $\endgroup$ – Hanno Apr 4 '17 at 20:40
  • $\begingroup$ @ Hanno, I also learn a fact non-trivial divisible groups are not finitely generated and $\mathbb{Z}_{p^\infty}$ is divisible. This can also explian $\mathbb{Q}$ is not finitely generated. $\endgroup$ – Daisy Apr 5 '17 at 2:11
  • $\begingroup$ @ Hanno, thank you very much. I understand it by your help. $\endgroup$ – Daisy Apr 5 '17 at 2:12

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