Question #16739

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.

Expert's answer

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

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