What requirements could be asked (minimal) of a ring R, so that any module M on R which is injective must also be projective? Is this possible?


Yes, the necessary and sufficient conditions are well-known.

The following conditions are equivalent for a ring $R$:

  1. All injective right $R$ modules are projectve
  2. All projective right $R$ modules are injective
  3. The injective and projective right $R$ modules coincide
  4. All of the above 3 conditions with "right" replaced with "left"
  5. $R$ Noetherian on a side and self-injective on a side
  6. $R$ is Artinian on both sides and self-injective on both sides
  7. $R$ is quasi-Frobenius

Any semisimple artinian Ring, will do the trick infact such a ring may be characterised as: Every R-Module is both projective and injective.

  • 1
    $\begingroup$ Just to be clear, semisimple artinian rings are characterized this way, yes, but not rings for which "injectives are projectives." $\endgroup$
    – rschwieb
    Feb 20 '14 at 20:42

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.