# Answer on Abstract Algebra Question for Melvin Henriksen

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.

(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)

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

*⇒*(2). Under (1), (*R/A*)*is a simple module, so Schur’s Lemma imply that the eigenring*_{R}*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.Need a fast expert's response?

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