This is a variation of Problem 8.21 on p.119 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition).
Let $K$ be a field, $R$ a subring of $K$ (with $1\in R$), and $I\ne\{0\}$ an ideal of $R$. Suppose $\beta\in K$ is such that for all $\alpha\in I$, we have $\beta\alpha\in I$. Is it true that $\beta\in R$?
In the original problem, $K$ is an algebraic-number field and $R$ the subring of algebraic integers; in that case $\beta\in R$ is true (though I can't prove it yet as I haven't finished reading the chapter on which this set of problems is based). I am just wondering if the problem can be meaningfully restated in the general context of fields and rings. Any ideas?
Thanks.
I'm sorry if this counterexample is too complicated, but I couldn't find a simpler one.
Let $K= \Bbb{Q}(X,Y)$, and $$R= \left\{ a+Y\frac{f(X,Y)}{g(X,Y)} : a \in \Bbb{Q}, f,g \in \Bbb{Q}[X,Y] , g(X,0) \neq 0 \right\} $$ You can check that $R$ is a (unital) subring of $K$ and that $X \notin R$. Call $$I=\left\{ a+Y\frac{f(X,Y)}{g(X,Y)} \in R: a = 0 \right\},$$ and $\beta=X$. Recall, $\beta \notin R$.
$I$ is an ideal of $R$, because it is the kernel of the ring morphism $$\begin{matrix} R &\longrightarrow & \Bbb{Q} \\ a + Y\frac{f(X,Y)}{g(X,Y)} & \longmapsto & a \end{matrix}$$
Finally, for all $\alpha = 0+ Y\frac{f(X,Y)}{g(X,Y)} \in I$ we have $$\beta \alpha = X \left( 0+ Y\frac{f(X,Y)}{g(X,Y)} \right) = 0+Y \frac{Xf(X,Y)}{g(X,Y)} \in I$$