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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

  

Source: Ágoston, István - Institute of Mathematics, Eötvös Loránd University

 

Collections: Mathematics