I am reading Daniel Marcus' book on Number Fields and I got stuck in the proof of Theorem $15$, Chapter $3$.
Theorem 15: Let $I$ be a nonzero ideal in a Dedekind domain $R$. Then there is a nonzero ideal $J$ such that $IJ$ is principal.
Before developing the proof, Marcus proves two lemmas:
Lemma 1: In a Dedekind domain, every nonzero ideal contains a product of nonzero prime ideals.
Lemma 2: Let $A$ be a nonzero proper ideal in a Dedekind domain $R$ with field of fractions $K$. Then there is an element $\gamma \in K\setminus R$ such that $\gamma A\subset R$.
Then the proof of Theorem 15 starts by
Let $\alpha$ be a nonzero member of $I$ and let $J=\left \{\beta \in R:\beta I\subset \left (\alpha\right )\right \}$. Then $J$ is easily seen to be an ideal (nonzero since $\alpha \in J$) and clearly $IJ\subset \left (\alpha\right )$. We will show that equality holds.
Consider the set $A=\frac{1}{\alpha}IJ$. This is contained in $R$ (recall $IJ\subset \left (\alpha\right )$), and in fact $A$ is an ideal (verify this). If $A=R$ then $IJ=\left (\alpha\right )$ and we are finished; otherwise $A$ is a proper ideal and we can apply Lemma 2. Thus $\gamma A\subset R$, $\gamma \in K\setminus R$. We will obtain a contradiction from this. Since $R$ is integrally closed in $K$, it is enough to show that $\gamma$ is a root of a monic polynomial over $R$.
Observe that $A=\frac{1}{\alpha}IJ$ contains $J$ since $\alpha \in I$; thus $\gamma J\subset \gamma A\subset R$. It follows that $\gamma J\subset J$; to see why this is true go back to the definition of $J$ and use the fact that $\gamma J$ and $\gamma A$ are both contained in $R$.
At this point I got stuck. I cannot see why $\gamma J\subset J$.
$\gamma A\subseteq R$ means $\frac{\gamma}{\alpha}IJ\subseteq R$, so for every $\beta\in J$, $\frac{\gamma\beta}{\alpha}I\subseteq R$, i.e. $\gamma\beta I\subseteq \alpha R=(\alpha)$. Since $\gamma\beta\in\gamma J\subseteq R$, this precisely means $\gamma\beta\in J$.