Let $R$ be a commutative ring. Let $S$ be a subset of $R$ and let $I$ be an ideal of $I$.
We define the subset
\[(I:S):=\{ a \in R \mid aS\subset I\}.\] Prove that $(I:S)$ is an ideal of $R$. This ideal is called the ideal quotient, or colon ideal.

Proof.

Let $a, b\in (I:S)$ and let $r\in R$. To show that $(I:S)$ is an ideal of the ring $R$, it suffices to show that the element $a+rb\in (I:S)$.
Thus we show that
\[(a+br)S\subset I.\]

Let $s\in S$ be an arbitrary element. Then since $a, b \in (I:S)$, we have $as, bs \in I$.
Since $I$ is an ideal of $R$, we have $r(bs)\in I$ as well.

Thus
\[(a+rb)s=as+r(bs)\in I\] for any $s\in S$, and hence we obtain $(a+br)S \subset I$.
By the definition of the ideal quotient, we have $a+br\in (I:S)$, and hence $(I:S)$ is an ideal of the ring $R$.

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