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Precisely, suppose $S$ is a submodule of the $R$-module $M$, then it is not necessarily possible to decompose $M$ as

$$ M = S \oplus S^c . $$

Any example?

It seems that there are many differences between vector spaces and modules. But I still cannot see the reason behind.

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kaiser
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    I assume $S^c$ is not the set complement of $S$, which is never a submodule. As an example, consider the $\Bbb Z$-module $\Bbb Z_4$ and the submodule $\Bbb Z_2$. This doesn't have a complement. – Pedro Oct 26 '15 at 15:40
  • Good example. But how to prove it? – kaiser Oct 26 '15 at 15:45
  • @kaiser Note that the existence of such a compliment would imply an isomorphism of $\mathbb{Z}$-modules $\mathbb{Z}_4\cong\mathbb{Z}_2\times\mathbb{Z}_2$. Can you see why this can't happen? – David Hill Oct 26 '15 at 15:47
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    @DavidHill Complement. – Pedro Oct 26 '15 at 15:59

1 Answers1

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Finding a direct-sum complement can fail in at least two ways:

  • $S$ could be an essential submodule (this means that for another submodule $T$, $S\cap T=\{0\}$ implies $T=\{0\}$)

  • $S$ could be a superfluous submodule (this means that for another submodule $T$, $S+T=M$ implies $T=M$.)

A good place to look for counterexamples is in rings which only have trivial idempotents, since idempotents correspond to module decompositions of the ring as a module over itself.

If you consider $\Bbb Z/4\Bbb Z$ as a module over itself, this module has exactly three submodules arranged in a line. The ring has no nontrivial idempotents. It's easy to see that the nontrivial submodule $2\Bbb Z/4\Bbb Z$ is both essential and superfluous at the same time, and so it (doubly) can't have a direct-summand complement.

Semisimple rings are exactly the class of rings for which you can always find direct-summand complements.

It seems that there are many differences between vector spaces and modules.

Of course. You might be interested in this question: Pathologies in module theory

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