 
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 EndZ 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 \ {b0} is a left ideal in R and it
is clearly minimal. Namely, let 0 = f I and g I be arbitrary elements. Then f(b0) = v = 0, hence there is r R
for which r(v) = g(b0). 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).
