 
Summary: Rings and Algebras Problem set #10: Solutions Nov. 24, 2011.
1. Let R be a semiperfect ring with a left ideal I. Suppose J(R) is nil. Show that I contains a nonzero idempotent
element whenever I is not contained in J(R).
Solution. Suppose I is not contained in J(R). Then I + J(R) is a left ideal in R/J(R) which is semisimple, hence it is generated by
an idempotent. Thus there is an x I such that x + I is idempotent. The proof of the statement about lifting idempotents modulo a
nil ideal show that the idempotent element e in the coset x + J(R) will be of the form xnp(x)n = p(x)n = xn hence it is contained in
the left ideal generated by x so it is also in I. (Note that the converse also holds since J(R) does not contain nonzero idempotents.)
2. Prove that an integral domain is semiperfect if and only if it is a local ring.
Solution. Clearly we have to show only one direction. Suppose R is an integral domain and semiperfect. If R/J(R) is not a field
then it must contain two nontrivial orthogonal idempotent elements which can be lifted orthogonally to R. This is impossible since
R is a domain.
3. Prove that a commutative ring is semiperfect if and only if it is a finite direct product of local rings.
Solution. Since R is semiperfect, RR = n
i=1 Rei for a set of orthogonal primitive idempotents and eiRei is local for each index i.
The commutativity implies that R is a ring direct sum of the local rings eiRei = Rei. (Note that the commutativity also implies that
the semisimple quotient R/J(R) is commutative, too, hence by the WedderburnArtin theorem it is a ring direct sum of commutative
fields. Thus R must be basic.) The converse is easy since the quotient modulo the radical is a finiet direct sum of fields, hence it is
semisimple; moreover in each summand of the factor ring there is only one nonzero idempotent which can be lifted to an idempotent
in the whole ring. Thus the sum of such idempotents in R/J(R) can also be lifted. Thus a finite direct product of commutative local
rings must be semiperfect.
