Lemma 10.28.1. Let $R$ be a ring. For a principal ideal $J \subset R$, and for any ideal $I \subset J$ we have $I = J (I : J)$.

**Proof.**
Say $J = (a)$. Then $(I : J) = (I : a)$. Since $I \subset J$ we see that any $y \in I$ is of the form $y = xa$ for some $x \in (I : a)$. Hence $I \subset J (I : J)$. Conversely, if $x \in (I : a)$, then $xJ = (xa) \subset I$, which proves the other inclusion.
$\square$

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