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?


  • 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 $q_0 + q_1 \alpha + q_2 \alpha^2 + \ldots + q_r \alpha^r=0$.
  • $\begingroup$ Note that you need $r_t = 1$ for $\beta$ to be an algebraic integer (and you have $q_r = 1$ in the condition on $\alpha$). $\endgroup$ – Magdiragdag Mar 15 '14 at 18:23

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.

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 $A$.


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.

  • 1
    $\begingroup$ Nice. I've seen a few questions on algebraic integers like "why is the sum of algebraic integers an algebraic integer?"; when I answered I always referred to the generic proof. This is a much more concrete approach; maybe much more appropriate if you haven't seen modules in general yet. Nevertheless, I think this only shows that poly's in algebraic int's are algebraic int's. The questions asks about transitivity of integrality. (Presumably, the question is not totally consistent about that). Can this argument be extended to show that a $\beta$ integral over $Z[\alpha]$ is integral itself? $\endgroup$ – Magdiragdag Mar 15 '14 at 18:28

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