248 D Eisenbud

morphism of left S-modules Hom^j(5, Q) is injective by hypothesis, so и sphts We will show that и is also essential (even as an Ä-homomorphism), thus proving that и IS an epimorphism

The condition that S be generated by finitely many elements in the central- izer of R IS easily seen to be equivalent to the condition that there exists an

n

epimorphism of Л-i^-bimodules jR"= Ц iR-^5 Using the fact that, for any

1=1

n

left R-module X, Homj,(i?",Z)s U ^ = ^" as left Ä-modules we obtain a

1=1

commutative diagram of left Ä-modules,

Homj , { S , Q)-----"-—. Hom;,(S, E)

Q - - - - - - - - - - - - - - - - - - - - - _^£п

where the vertical arrows are the monomorphisms induced by i?"->S Since ß-^E is essential, so is QJ^-^E^ It is easily seen that as submodules of E\ Hom^iS, Q) = ß"n Hom|j(S, E\ and thus и is essential Consider the commutative diagram

Hom^^ ( 5 , ß)-----î^—V Hom^(^, E)

Hom ; ^ ( Ä , ß) Hom;^(i^, E)

IW l\\

r\ _______inclusion rp

where the vertical arrows are induced by the inclusion R-^S Since E is tive, the vertical arrow on the right is an epimorphism We have already shown that и is an epimorphism, and it follows that ß -^ E is an epimorphism too Thus ß IS injective _j

Corollary 1. Let RC S be as in the theorem, and suppose that S is semismple Artinian Then R is semisimple Amman

Proof A ring IS semisimple Artmian iff each of its modules is injective _j Proof of Theorem i, a) By a theorem of Bass [3, Prop 4 1], it suffices to show that if {ß;fc}fcgj^ IS any set of mjective left Я-modules, thenUi^ß/t is again injective Of course, Нот^^(Х, ß^) is S-injective, and since S is Noethenan, \1k Hom^iCS, Qk) is S-injective too As S is a finitely generated left K-module,

Uk Hommes, ß,) = Hommes, Цкйк) An apphcation of Theorem 2 now finishes the proof