Answer to Question #16739 in Algebra for sanches

Question

For any right ideal A in a ring R, the idealizer of A is defined to be I(A) = {r ∈ R : rA ⊆ A}.
Show that I(A) is the largest subring of R that contains A as an ideal.

Answer 1

It is straightforward to check that IR(A) is a subring of R. Since A · A ⊆ A, A ⊆IR(A). Clearly A is an ideal in IR(A). Conversely, if A is an ideal in some subring S ⊆ R, then r ∈ S implies rA ⊆ A, so r ∈IR(A). This shows that S ⊆IR(A).