# Mahler measures of values of polynomials

Let $$K\ne \mathbb{Q}$$ be a number field, let $$\alpha\in \mathcal{O}_K$$ and let $$f(X)\in \mathcal{O}_K[X]$$. Denote the Mahler measure by $$M$$.

Is there any known result about the comparison of the values $$M(\alpha)$$ and $$M(f(\alpha))$$?

• I don't know, but it would be good to get some trivial cases out of the way. E.g., if $f$ is a constant polynomial, then clearly $M(f(\alpha))$ has nothing to do with $M(\alpha)$. – Gerry Myerson Apr 15 '19 at 5:27
• In which sense do you want to compare the values? Do you mean inequalities or something else? – François Brunault Apr 20 '19 at 8:51
• @FrançoisBrunault Yes, I mean inequalities. For example, do constants $C, \, n$ (depending on $K$ and $f$, but not on $\alpha$) exist such that $M(f(\alpha))\le C M(\alpha)^n$? A bound in the other direction is probably way more problematic. – Maurizio Moreschi Apr 20 '19 at 12:43

This is all standard stuff about height functions. More generally, if we use absolute heights, then for any $$f(x)\in\overline{\mathbb{Q}}[x]$$ of degree $$n$$ there are constants $$C_1(f)>0$$ and $$C_2(f)>0$$ so that for any $$\alpha\in\overline{\mathbb{Q}}$$, $$C_1(f)H(\alpha)^n \le H\bigl(f(\alpha)\bigr) \le C_2(f)H(\alpha)^n.$$ For $$C_2(f)$$ it's easy to get an explicit formula in terms of the coefficients of $$f$$ using the triangle inequality, as in Moreschi's answer. For $$C_1(f)$$, one can get an explicit formula that depends on the coefficients and the resultant of $$f$$. See for example Lang's Fundamentals of Diophantine Geometry.

Let $$H(\alpha)$$ denote the absolute multiplicative height of $$\alpha$$, which is a number in $$[1,\infty)$$, and $$M(\alpha) = H(\alpha)^d$$, where $$d$$ is the degree of $$\alpha$$. Basic properties of the height include:

• $$H(\alpha \beta) \leq H(\alpha)H(\beta)$$, for all algebraic numbers $$\alpha$$ and $$\beta$$,
• $$H(\alpha_1 + \cdots + \alpha_n) \leq n \prod_{i=1}^n H(\alpha_i)$$.

So if $$f(x) = c_n + c_{n-1}x + \cdots + c_0x^n,$$ we have $$H(f(\alpha)) \leq (n+1)\prod_{i=0}^n H(c_i) H(\alpha)^i \leq (n+1) \left[\max_{0 \leq i \leq n} H(c_i)\right]H(\alpha)^{n(n+1)/2}.$$ Let $$d$$ be the degree of $$\alpha$$, $$m = n(n+1)/2,$$ and let $$c = \max_{0 \leq i \leq n} H(c_i)$$. Raising both sides to the $$d$$-th power gives $$H(f(\alpha))^d \leq ((n+1)c)^d M(\alpha)^m.$$ Now $$f(\alpha) \in \mathbb{Q}(\alpha)$$, so the degree of $$f(\alpha)$$ is less than or equal to that of $$\alpha$$, and therefore $$M((f(\alpha))$$ is at most equal to the left-hand-side of the above inequality. [I had made a comment similar to this, but it contained an error, and so I deleted it.]

Note that $$d$$ is at most $$[K:\mathbb{Q}]$$, and so this upper bound is of the form $$CM(\alpha)^m$$, where $$m$$ depends only on $$f$$ and $$C$$ depends on $$K$$ and $$f$$.

As for the reverse direction, I don't think any such inequality can exist, because for an arbitrary $$\alpha$$ (i.e. of arbitrarily large Mahler measure), we may find a polynomial such that $$f(\alpha) = 1$$, so $$f(\alpha)$$ has Mahler measure as small as possible.

• The bound you prove can actually be improved to $H(f(\alpha))\le C H(\alpha)^n$ (which is what I was aiming for) under the assumption of my question. I post the proof of this as an answer, because it is too long for a comment. – Maurizio Moreschi May 6 '19 at 13:20
• There is a lower bound $H(f(\alpha))\ge CH(\alpha)^n$, but note that the implied constant will depend on $f$, as it will for the upper bound. – Joe Silverman May 6 '19 at 19:09

This the proof of the claim in my comment to Bobby Grizzard's answer.

Since, by definition $$H(\alpha):=\prod_{v\in \mathfrak{S}_K} \max\{1,|\alpha|_v\}\quad (\alpha\in K),$$ one has $$H(\alpha)=\prod_{v\in \mathfrak{S}_{K,\infty}} \max\{1,|\alpha|_v\}\quad \forall\alpha\in \mathcal{O}_{K},$$ where $$\mathfrak{S}_K$$, $$\mathfrak{S}_{K,\infty}$$ denote the set of all places of $$K$$ and the set of Archimedean places of $$K$$ respectively.

Now, let $$\alpha\in \mathcal{O}_{K}$$ and $$f(X)=a_n X^n+\dots+a_0\in \mathcal{O}_{K}[X]$$. Then $$f(\alpha)\in \mathcal{O}_K$$.

Moreover, for any $$v\in \mathfrak{S}_{K,\infty}$$ one has $$|f(\alpha)|_{v}\le f_v(|\alpha|_v),\qquad f_v(X):=|a_n|_{v} X^n+\dots+|a_0|_{v}\in \mathbb{R}_{\ge 0}[X].$$

It follows that $$|f(\alpha)|_{v}\le (|a_n|_{v}+\dots +|a_0|_{v}) \big(\max\{1,|\alpha|_v\}\big)^n$$ and thus $$\begin{split} H(f(\alpha))&=\prod_{v\in \mathfrak{S}_{K,\infty}} \max\{1,|f(\alpha)|_{v}\} \\ &\le \prod_{v\in \mathfrak{S}_{K,\infty}} \max\big\{1,(|a_n|_{v}+\dots +|a_0|_{v}) \big(\max\{1,|\alpha|_v\}\big)^n\big\} \\ & \le \prod_{v\in \mathfrak{S}_{K,\infty}} \max\{1,|a_n|_{v}+\dots +|a_0|_{v}\} \big(\max\{1,|\alpha|_v\}\big)^n \\ &= C_f H(\alpha)^n, \end{split}$$ where $$C_f:=\prod_{v\in \mathfrak{S}_{K,\infty}} \max\{1,|a_n|_{v}+\dots +|a_0|_{v}\}.$$