# Answer to Question #16739 in Algebra for sanches

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.

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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