On a step of Marcus' "Number Fields", Theorem 22 Chapter 3 I've been reading Marcus' Number Fields and there is a step in the proof of Theorem 22 that I can't understand. Specifically, it's about the proof of 22(b).

Theorem 22. Let $K$ and $L$ be number fields, $K\subset L$, let $R=\mathbb{A}\cap K$ (the set of algebraic integers in $K$) and $S=\mathbb{A}\cap L$, and $n=[L:K]$.
(a) For ideals $I$ and $J$ in $R$, $\lVert IJ\rVert=\lVert I\rVert\,\lVert J\rVert$.
(b) Let $I$ be an ideal in $R$. For the $S$-ideal $IS$, $\lVert IS\rVert = \lVert I\rVert^n$.
(c) Let $\alpha\in R$, $\alpha\neq 0$. For the principal ideal $(\alpha)$, $\lVert(\alpha)\rVert = |N_{\mathbb{Q}}^K(\alpha)|$.

Inside the proof of 22(b), Marcus states and proves (well, sort of) a Lemma.

Lemma: Let $A$ and $B$ be nonzero ideals in a Dedekind domain $R$, with $B\subset A$ and $A\neq B$. Then there exists $\gamma\in K$ such that $\gamma B\subset R$, $\gamma B\not\subset A$.

Then, he applies the Lemma with $A=P$ and $B=(b_1,\ldots,b_{n+1})$. Of course, we want a contradiction and so we assumed that $B$ is a subset of $A$. This is why we can apply the Lemma. But I am having trouble filling in the details. I can't get a contradiction.
Thanks a lot!
 A: I have a copy of Marcus with me, so I can fill in some of the details. However, unless you wanted the audience of your post to be limited to those who (i) have access to a copy of the book; and (ii) take the trouble to pull it off the shelf, look up the theorem, and backtrack enough to see what all the symbols mean; then you should try, in the future, to provide enough context in your post so that even those without access may have a shot at understanding the problem and your query about it, and potentially offer help.
Marcus will prove (b) in the special case of $I=P$ a prime ideal, relying on (a) to deduce the general case. Now, $S/PS$ is a vector space over $R/P$, and we want to show it has dimension exactly $n$. First, we show it has dimension at most $n$, by showing that any collection of $n+1$ elements is necessarily linearly dependent. To that end, let $\alpha_1,\ldots,\alpha_{n+1} \in S$, and we want to show that their images in $S/PS$ are linearly dependent over $R/P$. We know the original elements are linearly dependent in $L$ over $K$, since $[L:K]=n$. And we know that we can multiply the linear dependence equation by some integer so that all coefficients lie in $R$, rather than in $K$ (this was proven in an exercise in the previous chapter: if $\alpha\in K$, then there exists $m\in\mathbb{Z}$ such that $m\alpha\in R$). This gives us an equation of the form
$$\beta_1\alpha_1+\cdots+\beta_{n+1}\alpha_{n+1} = 0,$$
where $\beta_i\in R$. We want to reduce this modulo $P$, but to prove that we don't have a trivial linear combination after reduction, we need to ensure that not all $\beta_j$ lie in $P$. This is where the Lemma comes in.
If at least one $\beta_j\notin P$, we are done. Assume, however, that we are unlucky enough to have all $\beta_i\in P$. This could happen: for example, maybe all of our original $\alpha_i$ are in $P$, and so we pick  all $\beta_j$ in $P$. So we aren't looking for a contradiction. Instead, we want to show we can tweak the linear dependence relation so that the resulting one does  not have all coefficients in $P$.
We apply the Lemma with
$A=P$, $B=(\beta_1,\ldots,\beta_{n+1})$, ideals in the Dedekind domain $R$, we have $B\subseteq A$, $A\neq R$ (since it is prime). Let $\gamma\in K$ be guaranteed by the Lemma, so that $\gamma B\subset R$ and $\gamma B\not\subset P$. Now take the original linear dependent equation and multiply through by $\gamma$:
$$0 = \gamma(\beta_1\alpha_1+\cdots+\beta_{n+1}\alpha_{n+1}) = (\gamma\beta_1)\alpha_1+\cdots+(\gamma\beta_{n+1})\alpha_{n+1}.$$
Now notice that since $\gamma B$ is generated by $\gamma\beta_1,\ldots,\gamma\beta_{n+1}$, it cannot be the case that all the new coefficients lie in $P$ (since $\gamma B\not\subset P$). Thus, we are now in the situation where not all coefficients lie in $P$, and so reducing modulo $P$ we obtain a nontrivial linear dependence relation between $\overline{\alpha_1},\ldots,\overline{\alpha_{n+1}}$, which is what we wanted to show. This proves that $S/PS$ has dimension at most $n$, as claimed.
