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Consider short exact sequence of modules:

$ 0 \to A \to B \to C \to 0 $

It is well-known that $B$ consists isomorphic copy of $A$ which is image of $A$ under injective map and $C$ is the same as $B/A$.

Now consider:

$ 0 \to Z \to Z \to Z/nZ \to 0 $, where the first map is $z \to nz$ and the second - canonical projection.

Doesn't it mean that $Z$ is isomorphic to its own submodule which is ideal $nZ$? And from general case we have $Z/Z = Z/nZ$, so what's the matter? What do I understand the wrong way?

2 Answers 2

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If you have a surjective homomorphism $g:B\rightarrow C$ the isomorphism theorem says that you have an isomorphism $$ C\simeq\frac{B}{{\rm ker}(g)}. $$ Thus in the event of an exact sequence $0\rightarrow A\stackrel f\rightarrow B\stackrel g\rightarrow C\rightarrow 0$ you know that ${\rm ker}(g)$ is a copy of $A$ inside $B$ but in fact the displayed formula above reads $$ C\simeq\frac{B}{{\rm Im}(f)} $$ and this makes precise the fact that "the quotient $B/A$ is isomorphic to $C$".

Your confusion comes from the fact that modules may contain properly submodules which are (abstractly) isomorphic to themselves, but they have to be distinguished according to context.

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    Thus it is OK that module is isomorphic to its own submodule? I didn't know it is possible. – 2017-01-01
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    @Vladislav, if you accepted that subset can be equipotent to the whole set, there is really nothing strange here. You are probably used to finite dimensional vector spaces. – 2017-01-01
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    @Vladislav: For varying $n\neq0$ the subgroups $n\Bbb Z$ of $\Bbb Z$ are all isomorphic to each other as abelian groups (i.e. $\Bbb Z$-modules) because they are all cyclic infinite, but they are actually nested into each other according to the divisibilities among the $n$. – 2017-01-01
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$\mathbf Z$ is inded isomorphic as $\mathbf Z$-module to the ideal $n\mathbf Z$. However, we would have $\mathbf Z/\mathbf Z\simeq \mathbf Z/n\mathbf Z$ only if the square diagram:

\begin{alignat}{3} &\!\begin{aligned}[t]\mathbf Z\\\parallel\end{aligned}\xrightarrow{\;{}\times n\;\;}\begin{aligned}[t]\mathbf Z\\\parallel\end{aligned}{}\\ &\mathbf Z\,\xrightarrow{\;\;\operatorname{id}\;\;}\,\mathbf Z \end{alignat} were commutative, which is not the case, of course.