TITLE: Growth of the "denominator" of powers of an algebraic number
QUESTION [19 upvotes]: Let $z \in \overline{\bf Q}$ be an algebraic number.  Define the "denominator" of z to be the least natural number $n$ such that $nz$ is an algebraic integer.
By a rather ad hoc argument (playing with the minimal polynomial of $z$ to compute high powers of $z$ in terms of low powers of $z$, and measuring the coefficients obtained p-adically), I can show the following:

If $z$ is an algebraic number that is not an algebraic integer, then the denominator of $z^m$ grows exponentially in $m$ in the limit $m \to \infty$ (thus there is $c>1$ such that the denominator is at least $c^m$ for all sufficiently large $m$).

This is obvious in the case when $z$ is rational, from the fundamental theorem of arithmetic, but I couldn't find a similarly quick proof in the general case; presumably, unique factorisation into prime ideals for a suitable number field is the key, but (somewhat to my embarrassment) my algebraic number theory is too rusty to figure out how to exploit this here.  So I am posing the question here to see if anyone can find an easy proof of this fact.

REPLY [24 votes]: There is some number field $K$ containing $z$, and there we have a prime factorization
$$
(z)=\frac{\mathfrak{p}_1\ldots\mathfrak{p}_a}{\mathfrak{q}_1\ldots\mathfrak{q_b}}
$$
of fractional ideals, where no $\mathfrak{p}_i$ is equal to a $\mathfrak{q}_j$. If $nz^m$ is an algebraic integer, then $(\mathfrak{q}_1\ldots\mathfrak{q_b})^m$ divides $(n)$ (as ideals of $\mathcal{O}_K$). This means
$$
N(\mathfrak{q}_1\ldots\mathfrak{q_b})^m|N(n)=n^{[K:\mathbb{Q}]},
$$
so that
$$
n\geq \left(N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]}\right)^m.
$$
We can take
$$
c=N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]}
$$