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Let $R$ be a commutative ring with $1_R\neq 0_R$ and $S$ a sub-ring of $R$. I want to show that $(S:R)=\{i\in S:~ ir\in S,~\forall r\in R\}$ is the largest ideal in $S$.

I have tried to show that $(S:R)$ is indeed an ideal in $R$ since for any $s\in (S:R)$ and $r\in R$, if $i\in S$, then $$i(sr)=(is)r=r(is)=(ri)s=(ir)s=i(rs)\in S.$$

However, I have failed to show that $(S:R)$ is the largest ideal in $S$.

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Supose $I$ is an ideal of $R$ such that $I\subseteq S$.

Let $x\in I$.

Since $I$ is an ideal of $R$, for all $r\in R$, we have $rx\in I$, but then, since $I \subseteq S$, we get $rx\in S$.

Thus, by definition of $(S{\;\colon\,}R)$, we have $x\in (S{\;\colon\,}R)$.

It follows that $I\subseteq (S{\;\colon\,}R)$, hence $(S{\;\colon\,}R)$ is the largest ideal of $R$, contained in $S$.

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