Roots of monic polynomial over a number ring

Solution 1:

Hint: $\alpha$ is integral over $A$ if and only if $A[\alpha]$ is a finitely generated $A$ module. Also note that any submodule of a finitely generated $\Bbb Z$ module (or more generally a module over any noetherian ring) is finitely generated.

This is enough to show more generally that if $A\subseteq B\subseteq C$, $B$ integral over $A$, and $C$ integral over $B$, then $C$ integral over $A$.

Solution 2:

Since you demand a polynomial, I give you one. :)
The following method I found in the book Lectures on the theory of algebraic integers by Hecke.
If $\alpha$ satisfies a polynomial equation $f(x)=0$ with $f=\sum_0^na_ix^{n-i}\in R[x]$ and $a_0=1$, then these $a_i$ are algebraic integers over $R$, i.e. there are monic polynomials in $g_i\in\mathbb Z[x]$ such that $g_i(a_i)=0$. Then let $\{a_{ij}\}$, for every $i=0,\cdots,n-1$, be the set of all roots of $g_i$.
Now we finally come to our polynomial:$$h(x):=\prod_{j_1,\cdots,j_n}(x^n+a_{1,j_1}x^{n-1}+\cdots+a_{n,j_n}).$$ By definition we find that $h(\alpha)=0$, and that the coefficients of $h$ are symmetric functions in $a_{ij}$, hence in $\mathbb Q\cap R=\mathbb Z.$ Thus $\alpha$ is an algebraic integer. Q.E.D.
Any inappropriate point waits for a localisation.