I came across the following question, by reading proof of a theorem on structure of ring of integers in a number field (ref. Number Fields, Second Edition, by Marcus, P. $40$, Theorem $14$.)
Theorem: If $R$ is ring of integers of a number field $K$, & $0\neq I$ an ideal of $R$, then $|R/I|<\infty$.
Proof: (1) Consider $\alpha\neq 0$ in $I$.
(2) Let $[K:\mathbb{Q}]=n$; $\sigma_1,\cdots,\sigma_n$ be all the embeddings of $K$ in $\mathbb{C}$. Let $m=N^K(\alpha)=\prod \sigma_i(\alpha)$.
(3) Then $m\in\mathbb{Z}$. Let $\beta$ be product of the $n-1$ other conjugates of $\alpha$, so that $m=\alpha\beta$.
(4) (Then Marcus says:) The other conjugates of $\alpha$ may not be in $R$, but their product $\beta$ is in $R$; this is because, $\beta=m/\alpha$ is in $K$ [this is clear] and it is an algebraic integer [why?]
So my question is why the product of other $n-1$ conjugates of $\alpha$ (i.e. conjugates different from $\alpha$) is an algebraic integer?
I know another proof of the Theorem, but I am trying to understand the simple assertion in above proof of Marcus.