10
$\begingroup$

Is there a classification of the commutative rings (with unit) such that each module over the ring is projective ?

$\endgroup$
1
  • 7
    $\begingroup$ These commutative rings are exactly the finite products of fields. As a bonus, all their modules are injective as well as projective. They are also exactly the commutative semi-simple rings, where semi-simple is explained ( without the commutativity hypothesis) in tetrapharmakon's answer. $\endgroup$ Apr 20, 2011 at 22:36

1 Answer 1

17
$\begingroup$

They're called "semisimple artinian" rings. Prove that a ring $R$ (no commutativity is required) is semisimple artinian iff (equivalently)

0) (definition is most books in Ring Theory) $R$ is right artinian and has no nonzero nilpotent right ideals.

1) Any right R-module is projective.

2) Any right R-module is injective.

3) Any simple right R-module is projective.

4.1) Any right R-module is semisimple

4.2) R is a semisimple right module over itself (if you want, $R_R$ equals its socle).

5) $R$ consists of the sum of (finitely many) right ideals.

$\endgroup$
4
  • $\begingroup$ Notice also that you can substitute any occurence of "right" with "left", thanks to the fact that some of the conditions are right-left symmetric. ;) $\endgroup$
    – fosco
    Apr 20, 2011 at 22:09
  • 1
    $\begingroup$ And such a ring is necessarily the product of a finite number of fields, yes? $\endgroup$ Apr 20, 2011 at 22:23
  • 3
    $\begingroup$ Yes: Wedderburn-Artin.(In the non-commutative case you must take finite products of matrix rings over skew-fields) $\endgroup$ Apr 20, 2011 at 22:40
  • 1
    $\begingroup$ I think in condition 5) "simple right ideals" was intended. $\endgroup$
    – rschwieb
    Dec 17, 2011 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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