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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 $q_0 + q_1 \alpha + q_2 \alpha^2 + \ldots + q_r \alpha^r=0$.
kristof2014
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  • Note that you need $r_t = 1$ for $\beta$ to be an algebraic integer (and you have $q_r = 1$ in the condition on $\alpha$). – Magdiragdag Mar 15 '14 at 18:23

2 Answers2

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Worth emphasis: we can give a simple more concrete (high-school level) proof of the key Theorem in the accepted answer as follows. First note that we 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 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 $\, \alpha {\bf v} = L{\bf v},\ \beta {\bf v} = M{\bf v},\,$ i.e. in the products $\,\alpha^i\beta^j\,$ that occur as entries in $\,\alpha\bf v\,$ and $\,\beta\bf v, \,$ use the (minimal) polynomials of $\,\alpha\,$ and $\,\beta\,$ to rewrite $\,\alpha^n\,$ and $\,\beta^k\,$ into polynomials in lower powers of $\,\alpha\,$ and $\,\beta.\,$ Therefore

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

Hence, 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.

Bill Dubuque
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    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? – Magdiragdag Mar 15 '14 at 18:28
  • For a concrete example see thsi answer. $\ \ $ – Bill Dubuque Mar 14 '24 at 23:06
  • @Magdiragdag This post was intended as a supplement to the proof in your answer. – Bill Dubuque Mar 15 '24 at 06:50
  • @BillDubuque: (+1) Nice; I had not seen this answer before. It is nicely related to the answer that you cited. – robjohn Mar 15 '24 at 07:08
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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 $A$.

Magdiragdag
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