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:



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> 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 $m$).
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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.



Growth of the "denominator" of powers of an algebraic number
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
$$
c=N(\mathfrak{q}\_1\ldots\mathfrak{q\_b})^{1/[K:\mathbb{Q}]}
$$


