The Mahler measure of a integral polynomial is an algebraic integer

Question

Let $f(X) = a_nX^n+\cdots+a_0 \in \mathbb{Z}[X]$ be a polynomial with roots $\alpha_1,\dots,\alpha_n$. The Mahler measure of $f$ is defined to be $$ M(f) = |a_n|\prod_{i=1}^{n}\max\{1,|\alpha_i|\}. $$

I want to show that $M(f)$ is an algebraic integer. Below are my observations.

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First of all I tried to calculate the Mahler measure of a random polynomial (using SageMath). Take $f(X)=2X^3+4X^2-3X+5$, then $M(f) \approx 5.32791392520611...$, or explicitly,

$$ M(f)= 2 \times {\left| -\frac{1}{6} \, \left(\frac{17}{4}\right)^{\frac{1}{3}} {\left(3 \, \sqrt{5} \sqrt{3} - 13\right)}^{\frac{1}{3}} {\left(-i \, \sqrt{3} + 1\right)} - \frac{\left(\frac{17}{4}\right)^{\frac{2}{3}} {\left(i \, \sqrt{3} + 1\right)}}{3 \, {\left(3 \, \sqrt{5} \sqrt{3} - 13\right)}^{\frac{1}{3}}} - \frac{2}{3} \right|} {\left| \frac{1}{3} \, \left(\frac{17}{4}\right)^{\frac{1}{3}} {\left(3 \, \sqrt{5} \sqrt{3} - 13\right)}^{\frac{1}{3}} + \frac{2 \, \left(\frac{17}{4}\right)^{\frac{2}{3}}}{3 \, {\left(3 \, \sqrt{5} \sqrt{3} - 13\right)}^{\frac{1}{3}}} - \frac{2}{3} \right|} $$

So pretty sure the minimal polynomial of $M(f)$ cannot be seen too easily. Indeed, had the Mahler measure been very easy to calculate, then the Lehmer's conjecture wouldn't have been a thing.

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I also expect that the minimal polynomial of $M(f)$ to arise from the relation of roots and coefficients (i.e. the formulae of Vièta). Indeed, if all roots of $f$ are outside the closed unit disk, then $M(f)$ is nothing but $|a_0|$, which is already an integer, let alone algebraic. However, we cannot expect that all roots of $f$ are outside the closed unit disk. Nevertheless, we can notice that the Mahler measure $M(f)$ is somewhat a symmetric function w.r.t. all roots of $f$. I wonder if we can continue from here.

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We also know that $M(f)=\exp\left(\int_0^1 \log|f(e^{2\pi i t})|dt\right)$ so if we know something about $\log|a_n|$ and $\log^+|\alpha_i|$ we may be able to solve this question as well.

Answer 1

Notice first that the Mahler measure is multiplicative, hence it suffices to consider an irreducible polynomial. Namely, fixing $\alpha \in \overline{\Bbb Q}$, and letting $f \in \Bbb Z[X]$ be the minimal integer (in general non monic) polynomial of $\alpha$, it suffices to consider $M(f)$ in this context. Then, according to Proposition 1.6.6 of Heights in Diophantine Geometry (Bombieri, Gubler): $$M(f) = H(\alpha)^{\deg \alpha}$$ where $H(\alpha)$ is the multiplicative height of $\alpha$. Let me recall what this is.

Let $K = \Bbb Q (\alpha)$, and set $P_K$ the set of finite places of $K$. Then, for $v \in P_K$ above a prime $p_v$, and $x$ in the completion $K_v$, set $$|x|_v = |N_{K_v/\Bbb Q_{p_v}}(x)|_{p_v}^{\frac{1}{[K:\Bbb Q]}} = |N_{K_v/\Bbb Q_{p_v}}(x)|_{p_v}^{\frac{1}{\deg \alpha}}.$$ Then, $H(\alpha)$ is defined as $$H(\alpha) = \prod_{v \in P_K} \max(1, |\alpha|_v)$$ and so $$H(\alpha)^{\deg \alpha} = \prod_{v \in P_K} \max(1, |N_{K_v/\Bbb Q_{p_v}}(x)|_{p_v}).$$ I claim that this product is an algebraic integer. Indeed, every term on the right hand side is of the form $p_v^{a/b}$ with $a \ge 0$ (because of the maximum) and $b \ge 1$, which is a root of the monic, integer coefficient polynomial $X^b-p^a$. This concludes the argument.

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There is a much easier proof to show that $|a_n|^{n-1}M(f)$ is an algebraic integer. Write: $$f(X)=\sum_{i=0}^n a_i X^i=a_n\prod_{j=1}^n(X-\alpha_j) \in \Bbb Z[X].$$ Multiply by $a_n^{n-1}$: $$f(X)=\sum_{i=0}^n a_ia_n^{n-1-i}(a_nX)^i=\prod_{j=1}^n((a_nX)-a_n\alpha_j).$$ Set: $$g(Y)=Y^n+\sum_{i=0}^{n-1} a_ia_n^{n-1-i}Y^i=\prod_{j=1}^n(Y-a_n\alpha_j)$$ which is a monic polynomial with integer coefficients, and roots $a_n\alpha_j$. This shows that the $a_n\alpha_j$ are algebraic integers. Moreover, the complex norm of an algebraic integer is again an algebraic integer. Hence $$|a_n|^{n-1}M(f)=\prod_{j=1}^{n} \max(|a_n|,|a_n \alpha_j|)$$ is a product of algebraic integers, so is an algebraic integer.