$\beta\alpha$ in ideal $I$ for all $\alpha\in I$ $\implies$ $\beta$ is algebraic integer

Question

This is my second question on Problem 8.21 on p.119 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition); my first question has been answered.

Let $R=\mathcal O_K$ be the ring of integers of an algebraic field $K=\mathbb Q(\theta)$ for some $\theta$ algebraic over $\mathbb Q$. If $I\ne\{0\}$ is an ideal of $R$ and $\beta\in K$ is such that $\beta\alpha\in I$ for all $\alpha\in I$, prove that $\beta\in R$.

It is important that $K$ and $R$ be so defined, otherwise the result won't be true, as the counterexamples in my first question show. By studying the counterexamples, I presume that what may be important here are principally generated ideals. The ideal $I$ need not be principal but (by Theorem 8.13 on p.109 of the book) there exists an ideal $J\ne\{0\}$ such that the product $IJ$ is principally generated by a rational integer, i.e. $IJ=(c)$ where $c\in\mathbb Z$. I suspect this may be useful in solving the problem.

Another notion which may be useful is, given an ideal $I$, the set defined as $I^{-1}=\{\gamma\in K:\gamma\alpha\in R\ \text{for all}\ \alpha\in I\}$. Thus $\beta\in I^{-1}$ in the problem. If $J$ is an ideal, $JI^{-1}$ is defined as the set of all finite sums of products of elements of $J$ and of $I^{-1}$; if $J\subseteq I$ then $JI^{-1}$ is an ideal.

I have been puzzling over this problem for a while and would welcome any help with it. Thanks.

Answer 1

We are given that $\beta I\subseteq I$. If $I$ is principal, say $I=(\iota)$ with $\iota\ne 0$, then $\beta \iota= r\iota $ for some $r\in R$ and hence $\beta=r\in R$.

For any other non-zero ideal $J$, we can replace $I$ with $IJ$: If $\beta I \subseteq I$ then also $\beta IJ\subseteq IJ$. Using Theorem 8.19 we can pick $J$ so that $IJ$ is a non-zero principal ideal and are done.

Answer 2

You are given the inclusion $ \beta I \subset I $, multiplying both sides by the fractional ideal $ I^{-1} $ gives $ (\beta) \subset \mathcal O_K $, in particular, $ \beta \in \mathcal O_K $.

Answer 3

You don't need to argue with principal ideals if you don't want to. Instead, note that $I$ is finitely generated over $R$, say by $x_1,\ldots,x_n$. Then $\alpha \cdot x_i = \sum_j r_{ij} x_j$ for some $r_{ij} \in R$ (by the assumption), and so the matrix $\bigl( \alpha\delta_{ij} - r_{ij} \bigr)$ annihilates $I$. Thus so does its determinant; since $I \neq 0,$ we conclude that this determinant equals zero.

This determinant is a monic polynomial with entries in $R$, and thus $\alpha$ satisfies a monic polynomial with algebraic integer coefficients, and so is itself an algebraic integer.