Question #16812

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.

Expert's answer

(2) *⇒*(1) is clear, since *S *= *R *under the assumption (2).

(1)*⇒*(2). Under (1), (*R/A*)_{R} is 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.

(1)

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