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edited May 3, 2019 at 2:26

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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.

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.]

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.

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.

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created May 3, 2019 at 2:18

1.5k
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21

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.]

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.