# Understanding a proof in Algebraic Number Theory

Lemma: Let $$I$$ be a proper ideal in the ring $$R$$ of all algebraic integers, and $$\alpha_1,\dots,\alpha_n$$ algebraic over $$\mathbb{Q}$$. Then there exists $$\beta\in K=\mathbb{Q}(\alpha_1,\dots,\alpha_n)$$ such that $$\beta\alpha_i\in R$$ and not all $$\beta\alpha_i$$ belong to $$I$$.

Proof: (1) Let $$R_K$$ be the ring of integers in $$\mathbb{Q}(\alpha_1,\dots,\alpha_n)$$; it is a Dedekind domain.

(2) Let $$A$$ be the $$R_K$$-module generated by $$\alpha_1,\dots,\alpha_n$$.

(3) Since $$R_K$$ is Dedekind, there is a finitely generated $$R_K$$-module $$B$$ inside of $$R_K$$ such that $$AB=R_K$$.

(4) Since $$AB\nsubseteq I$$, there is $$\beta\in B$$ such that $$\beta A\nsubseteq I$$.

I don't understand (3) and (4), can one explain a little bit.

I was also doing in following way: $$\alpha_1,\dots,\alpha_n$$ are algebraic numbers; we can certainly find an integer $$m$$ in $$\mathbb{Z}$$ such that $$m\alpha_i$$ are all algebraic integers. After this I couldn't proceed to complete proof.

(3) is false as stated, it should be changed to "there is a finitely generated $R_K$-module $B$ inside $K$". For a counterexample, take $K = \mathbf Q(i)$ and $A = (2i)$, for instance.
The modified version of (3) follows from the invertibility of ideals in Dedekind domains. Any finitely generated $R_K$ submodule of $K$ is a fractional ideal, and it has an inverse. Take $B$ to be the inverse of the fractional ideal $A$.
(4) follows because $I$ is a proper ideal of $R$, thus $1 \notin I$, for example. Since $AB = R_K$, we cannot have that $AB \subset I$ (because it contains $1$.)