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Let $A$ be a $\mathbb{Z}$-module and suppose that $\mathbb{Z}\oplus \mathbb{Z}\cong \mathbb{Z}\oplus A$. Do we have $\mathbb{Z}\cong A$?

I know the result is true if $A$ is finitely generated as $\mathbb{Z}$-module. Clearly, $A$ can be viewed as a submodule of $\mathbb{Z}\oplus \mathbb{Z}$.

Maybe a submodule of $\mathbb{Z}\oplus \mathbb{Z}$ is finitely generated?

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    $\begingroup$ Just a comment about your last question. The module $\mathbb{Z} \oplus \mathbb{Z}$ is noetherian (it's finitely generated over a noetherian ring). Hence any submodule of $\mathbb{Z} \oplus \mathbb{Z}$ is finitely generated $\endgroup$
    – leibnewtz
    Jun 27, 2020 at 22:05
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    $\begingroup$ In fact, it is more generally true that if $A$ and $B$ are abelian groups and $\mathbb{Z}\oplus A\cong\mathbb{Z}\oplus B$, then $A\cong B$. See, for example, this question, although the case $B=\mathbb{Z}$ that you ask about is easier than the general case. $\endgroup$ Jun 28, 2020 at 10:09

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If $\Bbb Z \oplus A$ is finitely generated, then $A$ must be finitely generated. So, it indeed holds that if $\Bbb Z \oplus \Bbb Z \cong \Bbb Z \oplus A$, then $A$ is finitely generated and your previous observation applies. So, $A \cong \Bbb Z$.

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    $\begingroup$ And why exactly is A finitely generated? In general submodules of finitely generated modules need not be finitely generated right? $\endgroup$
    – user745578
    Jun 27, 2020 at 22:05
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    $\begingroup$ To be explicit, if $(n_1,a_1),\cdots,(n_k,a_k)$ generate $\mathbb{Z}\oplus A$, then $a_1,\cdots,a_k$ generate $A$. $\endgroup$
    – tkf
    Jun 27, 2020 at 22:05
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    $\begingroup$ @user745578: Though subgroups of f.g. modules need not be f.g., quotients do; and $A$ is a quotient of $\mathbb{Z}\oplus A$. $\endgroup$ Jun 27, 2020 at 22:09
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    $\begingroup$ More generally, the if a f.g. module surjects onto another module, that module is also f.g. (its generators being the images of the original generators under the surjection). So, in particular, any direct summand of a f.g. module is f.g. by virtue of the projection map. $\endgroup$
    – Thorgott
    Jun 27, 2020 at 22:09

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