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)?

  • 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


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 $M$.
  • $\begingroup$ 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

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