Rings and Algebras Problem set #4: Solutions Sept. 6, 2011. 1. a) Let S # R, and suppose RR is semisimple. Does it follow that S S is also semisimple? Summary: Rings and Algebras Problem set #4: Solutions Sept. 6, 2011. 1. a) Let S # R, and suppose RR is semisimple. Does it follow that S S is also semisimple? b) Is it true that any ring S can be embedded into a semisimple ring R? Solution. a) No, take for example Z # Q. -- b) We know that any semisimple ring is artinian (and noetherian) since the identity element is contained in a finite direct sum of simple modules, hence RR is a finite direct sum. Now, if S = R 1 × R 2 × ˇ ˇ ˇ is the cartesian product of infinitely many (nontrivial) rings, then the identity elements of the components will give an inifinite set e 1 , e 2 , . . . of orthogonal idempotents in S and hence also in any ring R containing S. Then Re 1 # R(e 1 + e 2 ) # ˇ ˇ ˇ # R(e1+ e 2 + ˇ ˇ ˇ + en ) # ˇ ˇ ˇ is a strictly increasing sequence of left ideals in R, hence R is not noetherian. Thus S cannot be embedded into a semisimple ring. 2. Determine which of the following abelian groups are semisimple: Z, Q, Q/Z, Zn , Z 2 # Z 3 # Z 5 # ˇ ˇ ˇ , Z 2 × Z 3 × Z 5 × ˇ ˇ ˇ , Z 2 × Z 2 × Z 2 × ˇ ˇ ˇ . Solution. Observe that any submodule of a semisimple module is semisimple. (This does not contradict the fact seen in the previous problem that semisimple rings can have non­semisimle subrings.) For example if M is semisimple and N # M then N is also a direct summand, hence a homomorphic image of M . The images of the simple submodules of M will be simple and will generate N , hence N is also semisimple. Since Z Z is not semisimple -- it does not have any simple submodules -- semisimple abelian groups cannot have torsion­free elements. Thus we get that Z, Q and Z 2 × Z 3 × Z 5 × ˇ ˇ ˇ are not semisimple. Z 2 # Z 3 # Z 5 # ˇ ˇ ˇ is almost by definition semisimple, and similarly, Z 2 × Z2 × Z 2 × ˇ ˇ ˇ is semisimple since each of its nonzero elements is contained in a simple abelian group Z2 . The group Q/Z contains submodules isomorphic to Zp # which are not semisimple (they are not complemented) hence it is not semisimple. Finally, Zn is semisimple if and only if n is squarefree. 3. Let M be a semisimple module. a) Show that if M is a direct sum of isomorphic simple modules, say M = #S (such semisimple modules are Collections: Mathematics