Answer to Question #16812 in Abstract Algebra for Melvin Henriksen
S = IR(A) be the idealizer of A. Show that (1), (2) are equivalent:
(1) A is a maximal right ideal and R/A is a cyclic left S-module.
(2) A is an ideal of R, and R/A is a division ring.
(2) ⇒(1) is clear, since S = R under the assumption (2). (1) ⇒(2). Under (1), (R/A)Ris a simple module, so Schur’s Lemma imply that the eigenring E := S/A is a division ring. If S(R/A) is cyclic, R/A is a 1-dimensional left E-vector space. Since S/A is a nonzero E-subspace in R/A, we must have S = R, from which (2) follows immediately.