2
$\begingroup$

Prove that:

$R$ is a semisimple ring $\Longleftrightarrow$ Every right $R$-module is injective (projective)

My try: $R$ is semisimple ring $\Longleftrightarrow$ Every right $R$-module is semisimple $\Longleftrightarrow$ Every submodule is direct summand

Please explain that why since every submodule is direct summand then every $R$-module is injective (projective)?

$\endgroup$
  • 4
    $\begingroup$ If every submodule is a direct summand, then every short exact sequence splits. $\endgroup$ – Mariano Suárez-Álvarez Nov 14 '13 at 16:05
4
$\begingroup$

Hints:

Use these characterizations

  • $E$ is injective iff every short exact sequence $0\to E\to B\to C\to 0$ splits for all $B,C$
  • $P$ is projective iff every short exact sequence $0\to A \to B\to P\to 0$ splits for all $A,B$
  • if $N<M$, then $0\to N \to M\to M/N\to 0$ splits iff $N$ is a summand of $M$.
$\endgroup$
  • $\begingroup$ So can I use Corollary 13.10 of Rings and Modules A&F? and use proposition: $E$ is injective module $\Longleftrightarrow$ every monomorphism $\phi : E \longrightarrow B$ splits. $\endgroup$ – Rachel Nov 14 '13 at 16:38
  • $\begingroup$ Dear @Rachel It's just one idea. There are a lot of other approaches that would work too, but I'm not sure what you're comfortable with. $\endgroup$ – rschwieb Nov 14 '13 at 17:07
  • $\begingroup$ OK, thanks. Please see math.stackexchange.com/questions/555507/…, u answered it before. I updated my question! $\endgroup$ – Rachel Nov 14 '13 at 17:15

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.