algebraic integer $\alpha$ + polynomial relation $\beta$ and $\alpha$ $\Rightarrow$ $\beta$ algebraic integer. Assume $\beta$ can be expressed in terms of polynomial relation in $\mathbb{Z}[\alpha]$. Where $\alpha$ is an algebraic integer (i.e. $\alpha$ is the root of a polynomial in $\mathbb{Z}[X]$. 
How can I argue that $\beta$ is itself an algebraic integer?
Attempt:


*

*assume: $r_0 + r_1 \beta+ r_2 \beta^2 + \ldots + r_t \beta^t=0$  with $r_i \in \mathbb{Z}[\alpha]$

*assume wlog, that $r_1$ contains some exponent $\alpha^l$

*how can I substitue that exponent, since I only have the condition $q_0 + q_1 \alpha + q_2 \alpha^2 + \ldots + q_r \alpha^r=0$. 

 A: There is a way more concrete than the proof from Atiyah-MacDonald, that lends further insight.
Namely one can represent algebraic integers as eigenvalues of square-integer matrices. Such matrices have monic integral characteristic polynomials, so their eigenvalues are algebraic integers. Conversely, $\alpha\,$ is an eigenvector of its companion matrix, i.e. the matrix of the linear map $\,x\mapsto \alpha x\,$ on $\,\Bbb Z[1,\alpha,\ldots\,\alpha^{n-1}].\,$ In $\,\Bbb Z[1,\alpha,\ldots,\alpha^{n-1}]\!\times\! [1,\beta,\ldots,\beta^{k-1}]=\Bbb Z[\alpha^i\beta^j],\,$ $\, i<n,\,j<k,\,$ with basis $\,{\bf v} = [\alpha^i\beta^j]\,$ there are matrices $\,L,M\,$ such that $\, \alpha {\bf v} = L{\bf v},\ \beta {\bf v} = M{\bf v},\,$ therefore
$$\begin{eqnarray} 
(L+M){\bf v} &=& L{\bf v} + M{\bf v} = \alpha{\bf v} + \beta{\bf v} = (\alpha+\beta){\bf v}\\ \\
(LM){\bf v} &=& L(M{\bf v}) = L(\beta{\bf v}) = \beta(L{\bf v}) = \beta\alpha {\bf v}\\
\end{eqnarray}$$
Therefore, being eigenvalues of square integer matrices, $\,\alpha+\beta\,$ and $\,\alpha\beta\,$ are algebraic integers. Further details can be found e.g. in Robin Chapman's Notes on Algebraic integers, pp. 2-3.
Since algebraic integes are closed under sums and products, the sought result follows immediately, since polynomials $\in\Bbb Z[\alpha]\,$ are compositions of sums and products of algebraic integers.
A: The standard argument for this requires a bit of machinery, namely the following theorem (the formulation below is from Atiyah-MacDonald, Proposition 5.1).
Theorem. Let $B$ be a (commutative) ring and $A$ a subring and $x \in B$. Then the following four statements are equivalent.


*

*$x$ is integral over $B$;

*$A[x]$ is a finitely generated $A$-module;

*$A[x]$ is contained in a subring $C$ of $B$ such that $C$ is a finitely generated $A$-module;

*There exists a faithful $A[x]$-module $M$ which is finitely generated as an $A$-module.


Then, as a corollary (Corollary 5.4. in Atiyah-MacDonald), you get the following.
Corrolary. Let $A \subseteq B \subseteq C$ be (commutative) rings. Assume that $B$ is integral over $A$ and that $C$ is integral over $B$. Then $C$ is integral over $A$.
Proof Take $x \in C$. Because $C$ is integral over $B$,
$$x^n + b_{n-1} x^{n-1} + \dots + b_0 = 0$$
for certain $b_0, \dots, b_{n-1} \in B$. The ring $B' = A[b_1,\dots,b_n]$ is a finitely generated $A$-module (repeatedly the theorem above) and $B'[x]$ is a finitely generated $B'$-module (also by the theorem above). Hence $B'[x]$ is a finitely generated $A$-module and therefore (again by the theorem above) $x$ is integral over $A$.
