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

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    $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Apr 20 '11 at 22:36
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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.

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  • $\begingroup$ 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 Loregian Apr 20 '11 at 22:09
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    $\begingroup$ And such a ring is necessarily the product of a finite number of fields, yes? $\endgroup$ – Tom Goodwillie Apr 20 '11 at 22:23
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    $\begingroup$ Yes: Wedderburn-Artin.(In the non-commutative case you must take finite products of matrix rings over skew-fields) $\endgroup$ – Georges Elencwajg Apr 20 '11 at 22:40
  • $\begingroup$ I think in condition 5) "simple right ideals" was intended. $\endgroup$ – rschwieb Dec 17 '11 at 18:37

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