Rings and Algebras Problem set #2: Solutions Sept. 22, 2011. 1. Show that the converse of Schur's lemma does not hold. Summary: Rings and Algebras Problem set #2: Solutions Sept. 22, 2011. 1. Show that the converse of Schur's lemma does not hold. Solution. Consider the additive group of Q as a module over Z. Then Q is clearly not simple but End Z Q = Q, since each endomorphism # is determined by the image #(1) = # # Q and it is also clear that multiplication by # # Q gives an endomorphism. 2. Let V be a vector space over a field K. Show that R = End(VK ) is left primitive but not necessarily simple. Describe the ideal structure of R. Solution. Clearly V as a left R­module is faithful and simple. Hence R is left primitive. If dimK V is finite, then R is simple, so let us consider the infinite dimensional case. It is easy to show that if # is an arbitrary infinite cardinality then the endomorphisms of rank less than # form and ideal I# . Furthermore if f # R is of rank # then any g # R of rank not greater than # will belong to the ideal generated by f . Hence the ideals are all of this form. Thus the ideals of R form a chain and they are in one­to­one correspondence with infinite cardinalities # such that # # (dim V ) + , the correspondence being given by # ## I# . 3. Show that if R = End(VK ) as above then R has a minimal left ideal and conclude that every simple faithful left R­module is isomorphic to RV . Solution. Let dim V = # and B = {b # | # < #} be a basis for V . Then I = AnnR B \ {b 0 } is a left ideal in R and it is clearly minimal. Namely, let 0 #= f # I and g # I be arbitrary elements. Then f(b 0 ) = v #= 0, hence there is r # R for which r(v) = g(b 0 ). With this r we have rf = g, hence Rf = I. This implies that I is minimal. Since any primitive ring is prime, a statement from the lecture implies that every faithul simple R­module is isomorphic, hence they must be isomorphic to RV (which is faithful and simple). Collections: Mathematics