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I have read that a module is module is noetherian if and only if every submodule is finitely generated. And I have proved it using the axiom of choice.

However, I came across this problem recently and found that this somehow looks like a counterexample.

let $\mathbb{Z_{p^{\infty}}}$denote the submodule of the $\mathbb{Z}$-module $\mathbb{Q/Z}$ consisting of elements which are annihilated by some power of p. Show that $\mathbb{Z_{p^{\infty}}}$is an artinian $\mathbb{Z}$-module which is not noetherian.

I proved this problem by proving every submodule of $\mathbb{Z_{p^{\infty}}}$ must have a smallest ${\frac{1}{p^{k}}}$,but this somehow hints that every submodule is finitely generated(by ${\frac{1}{p^{k}}}$), which looks quite contradicting with the proposition above.

Can anybody help explain what is wrong here?

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Notice that $\mathbb{Z} _{p^{\infty}}$ itself is not finitely generated.

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  • $\begingroup$ but every submodule is.the propostitionsaid that if and only if every submodule is finitly generate. $\endgroup$ – Alex Jan 7 '13 at 16:13
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    $\begingroup$ Every proper submodule of your module is finitely generated, but in the definition of being Neotherian we require every submodule to be finitely generated, including the whole module itself. $\endgroup$ – Piotr Pstrągowski Jan 7 '13 at 16:14

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