I have trouble understanding the outlined argument. We can multiply each $\mu_i$ by $a_i^{n-1}$ to make it integral, where $a_i$ is a leading coefficient of the minimal polynomial of $\mu_i$, but how can we make a single multiplier for all of them?
Note. It may not be clear from the context that:
- $L,K$ - fields, $L|K$ - finite;
- $K = Quot(A)$;
- $B$ - integral closure of $A$ in $L$.
Edit: I am now also confused by the end of this proof:
Isn't $adM$ a submodule of $M$?
Using the same notation as in your question.
If $\alpha_i\in L$ is algebraic over $K$ then there is $a_i\in A$ such that $a_i\alpha_i $ is integral over $A$ (one only needs to write a monic equation for $\alpha_i$ to see that). This means $a_i\alpha_i \in B$.
Now suppose we have $\alpha_1,\ldots, \alpha_k \in L$ all algebraic over $K$, then we can take $a=a_1\cdots a_k$ so that $d\alpha_i$ is integral over $A $ for all $i$ (you only need to remember that $B$ is a ring). That is $a\alpha_i \in B$ for all $i$.
For why $M$ is a free $A-$module if $\lambda M$ is a free $A-$module for nonzero $a\in A$, one needs to remember that everything is happening inside $L$, which is a field, so in particular $M$ is torsion-free over $A$. This will guarantee that if $\{\lambda \beta_i\}_i$ is a basis of $\lambda M$ then $\{\beta_i\}_i$ is a basis of $M$. In particular, they have the same rank.