0
$\begingroup$

I have to demonstrate that $R$ viewed as left $R$-module is a progenerator (i.e. a projective and finitely generated generator) of $R$-mod, the category of left $R$-modules.

How can I do this? I know that a left $R$-module $M$ is a generator of $R$-mod if and only if $M^n$ is the direct sum of $R$ and $R'$ for some integer $n>1$ and some $R$-module $R'$. Thanks!

$\endgroup$
  • $\begingroup$ $R^2 \cong R \oplus R?$ $\endgroup$ – Aurel Aug 14 '18 at 12:49
  • $\begingroup$ Do you know what "projective" and "generator" means? $\endgroup$ – Arnaud D. Aug 14 '18 at 12:53
  • $\begingroup$ The definition I have found is that a left $R$-module is a generator if it generates every module in $R$-mod and that a $P$ module is projective if, given any epimorphism $p:M \to N$, every morphism $f:P \to N$ can be factorized as $f=pg$ for some $g:P \to M$. $\endgroup$ – robbis Aug 14 '18 at 13:28
  • $\begingroup$ @Aurel so this is ovious to say that $R$ is a generator $\endgroup$ – robbis Aug 14 '18 at 13:32
  • $\begingroup$ According to the characterization you gave of a generator in your post, I would say that it is. $\endgroup$ – Aurel Aug 14 '18 at 13:51

Your Answer

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

Browse other questions tagged or ask your own question.