1
$\begingroup$

M is a semisimple module iff every submodule of M is a direct summand (here is the definition of semisimple modules, and this is the property 3), but this property can be replaced with "every submodule of M is $\textit{isomorphic}$ to a direct summand of M"?

$\endgroup$
  • 2
    $\begingroup$ Of course not. For example, every submodule of $\Bbb{Z}$ is isomorphic either to $\Bbb{Z}$ or $0$, which are direct summands of $\Bbb{Z}$; however $\Bbb{Z}$ is not a semisimple $\Bbb{Z}$-module. $\endgroup$ – Crostul Nov 20 '16 at 21:48
  • $\begingroup$ Thank you, that is a nice example $\endgroup$ – adiselann Nov 20 '16 at 22:18
2
$\begingroup$

Of course not. For example, every submodule of $\Bbb{Z}$ is isomorphic either to $\Bbb{Z}$ or $0$, which are direct summands of $\Bbb{Z}$, however $\Bbb{Z}$ is not a semisimple $\Bbb{Z}$-module.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.