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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.

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$I_A(J)$ 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 $S \subset I_A(J)$.

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