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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.

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  • $\begingroup$ The point should be that integrality is a local condition ($z$ is an algebraic integer iff $\nu(z) \ge 0$ for every valuation $\nu$ on $\mathbb{Q}(z)$) and I feel like there should be a clean proof of this but I don't know it. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 22, 2013 at 1:20
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    $\begingroup$ @QiaochuYuan: Take the ideal $I$ of algebraic integers $a\ in \mathbb Q(z)$ such that $az$ is an integer. $I$ is a proper nontrivial ideal, so it has some proper nontrivial prime factor. This defines a valuation. The valuation of $z$ by that valuation cannot be nonnegative, or else $z$ could be written as $b/a$ for $b$ an algebraic integer and $a$ a unit mod the prime ideal, which is a contradiction because $a$ is not in $I$. $\endgroup$
    – Will Sawin
    Commented Nov 22, 2013 at 3:02

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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 $$ c=N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]} $$

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    $\begingroup$ Ah, that was indeed simple - somehow I had not figured out that one should be taking norms :-) $\endgroup$
    – Terry Tao
    Commented Nov 22, 2013 at 1:55

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