Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a ring and let $J$ be a right-sided ideal of $A$. We call the set $I_{A}(J)=\lbrace a \in A \mid aJ\subset J\rbrace$ the idealizer of $J$.

Show that $I_{A}(J)$ is the largest subring of that $A$ containing $J$ as an ideal.

share|improve this question

1 Answer 1

$I_A(J)$ is a subring of $A$:

Let $s,r\in I_A(J)$. Then $(r+s)J = rJ + sJ \subset J$ and $(rs)J = r\cdot sJ \subset rJ \subset J$, so $I_A(J)$ is closed under addition and multiplication.

$I_A(J)$ contains $J$ as a two-sided ideal:

This follows directly from the definition of $I_A(J)$.

$I_A(J)$ is the largest subring of $A$ containing $J$ as a two-sided ideal:

Let $S$ be a subring of $R$ such that $I_A(J)$ is a two-sided ideal in $S$. Then for all $s\in S$, $sJ \subset J$, showing that $S \subset I_A(J)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.