algebraic integer $\alpha$ + polynomial relation $\beta$ and $\alpha$ $\Rightarrow$ $\beta$ algebraic integer.

Question

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

Answer 1

Worth emphasis: we can give a simple more concrete (high-school level) proof of the key Theorem in the accepted answer as follows. Recall we can represent algebraic integers as eigenvalues of square-integer matrices. Such matrices have integral characteristic polynomials $\in\Bbb Z[x]$ that are monic (lead coef $ = 1)$ so their eigenvalues are algebraic integers. Conversely, $\alpha\,$ is an eigenvalue of its companion matrix, i.e. the matrix of the linear map $\,x\mapsto \alpha \:\!x\,$ on $\,\Bbb Z[1,\alpha,\ldots\,\alpha^{n-1}].\,$

Now 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 $\,{\bf v} = [\alpha^i\beta^j],\,$ we can construct $\,nk \times nk\,$ matrices $\,L,M\,$ such that $\, \alpha\:\! {\bf v} = L\:\!{\bf v},\ \beta\:\! {\bf v} = M\:\!{\bf v},\,$ since in products $\,\alpha^i\beta^j\,$ that occur as entries in $\,\alpha\:\!{\bf v},\:\! \beta\:\!{\bf v}\,$ we can use the (minimal) polynomials of $\,\alpha\,$ and $\,\beta\,$ to rewrite $\,\alpha^n\,$ and $\,\beta^k\,$ into polynomials in lower powers of $\,\alpha\,$ and $\,\beta.\,$ Thus we get

$$\begin{eqnarray} (L+M)\:\!{\bf v} &=& L{\bf v} + M{\bf v} = \alpha\:\!{\bf v} + \beta\:\!{\bf v} = (\alpha+\beta){\bf v}\\[.5em] (LM)\:\!{\bf v} &=& L(M{\bf v}) = L(\beta{\bf v}) = \beta(L{\bf v}) = \beta\alpha\:\! {\bf v}\\ \end{eqnarray}\qquad$$

Thus, being eigenvalues of square integer matrices, $\,\alpha+\beta\,$ and $\,\alpha\beta\,$ are algebraic integers.

Algebraic integers are closed under sums and products, so the sought result follows immediately, since polynomials $\in\Bbb Z[\alpha]\,$ are compositions of sums and products of algebraic integers.

Example $ $ For a concrete calculation using this method see Robjohn's $2020$ answer here.

Readers familiar with linear algebra may find it instructive to express the above in the language of Kronecker products, e.g. see this answer, or The Arithmetic of Algebraic Numbers: An Elementary Approach, College Math Jnl, $2004$ (non-paywalled here).

Answer 2

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.

1. $x$ is integral over $B$;
2. $A[x]$ is a finitely generated $A$-module;
3. $A[x]$ is contained in a subring $C$ of $B$ such that $C$ is a finitely generated $A$-module;
4. 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.

Corollary. 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 $A$.