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Let $a$ and $b$ algebraic numbers over $\mathbb{Q}$. Do you know (or recall) if there are simple uppper bounds relating the logarithmic height of $ab$ (or $a/b$) with the logarithmic height of $a$ and the logarithmic height of $b$?
With logarithmic height of $a$ I mean here the logarithm of the absolute value of the largest (in the sense of absolute value) coefficient of the minimal polynomial of $a$.
Thanks for your answers.
Edit 1: Extended the argument for algebraic numbers.
Edit 2: Completed answer, finishing up the remaining case for $[\mathbb Q(\alpha\beta): \mathbb Q] < [\mathbb Q(\alpha):\mathbb Q][\mathbb Q(\beta):\mathbb Q]$. See section 3.
(Upper bound increased a little, might have been possible to keep it the same.)
Edit 3: Created a SageMathCell link to generate random examples.
Let $H(f)$ denote the (non-logarithm, absolute) height of a polynomial $f(x)$ and $h(f)$ the logarithmic version. i.e. if $f(x) = f_dx^d + f_{d-1}x^{d-1}+\cdots + f_0$ then $$ \begin{align} H(f) &:= \max_{0\leq i \leq \deg(f)}\{|f_i|\}\\ h(f) &:= \log(H(f)) \end{align} $$
Here is an upper bound which holds for most cases:
Proposition. Let $\alpha,\beta$ algebraic numbers, then $\alpha\beta$ is also an algebraic number. If $$[\mathbb Q(\alpha\beta): \mathbb Q] = [\mathbb Q(\alpha):\mathbb Q][\mathbb Q(\beta): \mathbb Q],$$ then it can be shown that $$ h(f_{\alpha\beta}) \leq \deg(f_\beta) h(f_\alpha) + \deg(f_\alpha)h(f_\beta) + C(\deg(f_\alpha),\deg(f_\beta)) $$ where $C(M,N)$ is a constant depending on $M,N$: $$ C(M,N) = \log(\binom{MN}{\lfloor MN/2\rfloor} (M+1)^{N/2} (N+1)^{M/2}) $$ Edit 2: For the remaining case, $$[\mathbb Q(\alpha\beta): \mathbb Q] < [\mathbb Q(\alpha):\mathbb Q][\mathbb Q(\beta): \mathbb Q],$$ the same equation holds for by replacing $C(\deg(f_\alpha),\deg(f_\beta))$ with a slightly larger $D(\deg(f_\alpha),\deg(f_\beta))$, see section 3. $$ D(M,N) = \log(\binom{MN}{\lfloor MN/2\rfloor}^2 \sqrt{MN+1} (M+1)^{N/2} (N+1)^{M/2}) $$
This is done by first obtaining an explicit form of $f_{\alpha\beta}(x)$, the minimal polynomial of $\alpha\beta$, then bounding the coefficients using Mahler measure.
This is suspected to holds for all $\alpha\beta$ without the degree restriction, but I have yet to find a way to show that. Edit 2: Done in section 3.
In practice it was observed that
$$
h(f_{\alpha\beta}) \approx \deg(f_\beta) h(f_\alpha) + \deg(f_\alpha)h(f_\beta)
$$
and examples where LHS is larger are also found. Hence we cannot drop the $C(M,N)$ term, but it might be possible to reduce it more. (Although as $h(f_{\alpha\beta})\to \infty$ the $C(M,N)$ becomes relatively small/insignificant.) A "best case" might be to reduce $C(M,N)$ to a constant $K$ that does not depend on anything.
Edit 2: same goes for the other case of $D(M,N)$.
We also remark that $C(M,N)$ turns out to be very close to $MN$.
Our first goal is to get an explicit form of $f_{\alpha\beta}(x)$, the minimal polynomial of $\alpha\beta$. By showing $$ t\cdot f_{\alpha\beta}(x) = g(x) $$ for some $t\in\mathbb Z$ and $g(x)\in\mathbb Z[x]$, we get an upper bound $$ H(g) = H(t\cdot f_{\alpha\beta}) = |t| H(f_{\alpha\beta}) \geq H(f_{\alpha\beta}) $$
Lemma: Let the minimal polynomials of $\alpha,\beta$ be $$ \begin{align} f_\alpha(x) &= a\prod_{i=1}^m(x-\alpha_i)\\ f_\beta(x) &= b\prod_{j=1}^n (x-\beta_j) \end{align} $$ respectively, where $(\alpha,\beta)=(\alpha_1,\beta_1)$ and $\{\alpha_i,\beta_j\}$ are the conjugates. Then $\alpha\beta$ is a root of a polynomial $g(x)\in\mathbb Z[x]$, given by $$ \begin{align} g(x) &:= a^nb^m\prod_{j=1}^n\prod_{i=1}^m (x-\alpha_i\beta_j) \end{align} $$ In particular, this means $$ g(x) = t\cdot f_{\alpha\beta}(x) $$ for some $t\in \mathbb Z$.
Proof. Using the resultant function, it can be shown that $\alpha\beta$ is a root of $$ g(x):= Res_y(f_\alpha(y), y^n f_\beta(x/y)), $$ as described in the Number Theory section.
Since $f_\beta(x)$ is degree $n$ in $x$ with integer coefficients, $y^nf_\beta(x/y)$ is homogenous degree $n$ in $x$ and $y$ with integer coefficients. Therefore the resultant is taken between two polynomials in $y$ with coefficients in $R =\mathbb Z[x]$. Since the resultant takes the determinant of a matrix in $R$, this shows that the result $g(x)$ is a polynomial in $\mathbb Z[x]$.
Using the two properties of resultant: $$ \begin{align} Res_y(\prod_{i=1}^m y-\alpha_i,\prod_{j=1}^n y - \beta_j) &= \prod_{j=1}^n\prod_{i=1}^m \alpha_i - \beta_j\\ Res_y(a f(y), b g(y)) &= a^{\deg(g)}b^{\deg(f)} Res_y(f(y),g(y)), \end{align} $$ this can be expanded to $$ \begin{align} g(x) &:= Res_y(a\prod_{i=1}^m y - \alpha_i, b\prod_{j=1}^n x - \beta_j y)\\ &= Res_y(a\prod_{i=1}^m y -\alpha_i, b(-\beta_j)^n \prod_{j=1}^n y - x/\beta_j)\\ &= a^n b^m(-\beta_j)^{mn} Res_y(\prod_{i=1}^m y -\alpha_i, \prod_{j=1}^n y - x/\beta_j)\\ &= a^n b^m(-\beta_j)^{mn} \prod_{j=1}^n\prod_{i=1}^m \alpha_i- x/\beta_j)\\ &= a^n b^m \prod_{j=1}^n\prod_{i=1}^m x-\alpha_i\beta_j \end{align} $$ Since $g(x)$ contains the factor $x-\alpha_1\beta_1 = x-\alpha\beta$, this shows that $g(\alpha\beta) = 0$, concluding the first part of the proof.
From field theory, $f_{\alpha\beta}(x)$ being a minimal polynomial means it divides $g(x)$ in $\mathbb Q[x]$, so $f_{\alpha\beta}(x)$ and $g(x)$ both having degree $mn$ means $$ t \cdot f_{\alpha\beta}(x) = g(x) $$ for some $t\in\mathbb Q$. If $t\not\in\mathbb Z$, $t=u/v$ for some $\gcd(u,v)=1$ and $u,v\in\mathbb Z$. Then from $\gcd(u,v)=1$, $$(v/u) f_{\alpha\beta}(x)=g(x)\in\mathbb Z[x] \implies (1/u)f_{\alpha\beta}(x)\in\mathbb Z[x],$$ so we find a smaller integral polynomial with root $\alpha\beta$, contradicting minimality of $f_{\alpha\beta}(x)$. Therefore $t\in\mathbb Z[x]$ and this completes the proof. $$\tag*{$\square$}$$
As mentioned at the start, this gives us a bound of $$ H(f_{\alpha\beta}) \leq H\left(a^nb^m\prod_{j=1}^n\prod_{i=1}^m (x-\alpha_i\beta_j)\right) $$ Remark 1: For algebraic integers, $(a,b)=(1,1)$ and the inequality becomes an equality.
Remark 2: For the other special case where $\alpha\beta$ has degree $<mn$, we can still show that $g(x)\in\mathbb Z[x]$ and therefore $$ g(x) = f_{\alpha\beta}(x) q(x), $$ for some $q(z)\in\mathbb Z[x]$. The problem here is there are no obvious ways to (upper-) bound $H(f_{\alpha\beta})$ using $H(g)$. For general polynomials $t(x) = r(x)s(x)$ we can have $H(r) > H(t)$ (Take $(x^2+10x+1)(x^2-x+1)$). The requirement that $f_{\alpha\beta}(x)$ has degree $mn$ resolves this problem.
Section 1 gave us the bound $$ H(f_{\alpha\beta}) \leq H(g), $$ we shall show that that $H(g)$ can in turn be bounded by $H(f_{\alpha})$ and $H(f_{\beta})$ using Mahler measure.
Lemma. Let $f_{\alpha}(x),f_{\beta}(x)$ be polynomials of the form $$ \begin{align} f_{\alpha}(x) &= a\prod_{i=1}^m x - \alpha_i\\ f_{\beta}(x) &= b\prod_{j=1}^n x - \beta_j\\ \end{align} $$ Then we have the height inequality $$ H(a^nb^m\prod_{j=1}^m\prod_{i=1}^n x-\alpha_i\beta_j) \leq H(f_{\alpha})^n H(f_{\beta})^m \binom{mn}{\lfloor mn/2 \rfloor}(m+1)^{n/2} (n+1)^{m/2} $$
Proof. The Mahler measure $M(p)$ of a polynomial $p(x)$ $$ p(x)= p_d\prod_{i=1}^d (x-\gamma_i) $$ is defined as $$ M(p) = |p_d|\prod_{i=1}^d \max\{1,|\gamma_i|\} $$ i.e. (absolute) product of leading coefficient and roots with absolute value $>1$.
It satisfies the inequality $$ \binom{d}{\lfloor d/2\rfloor}^{-1} H(p) \leq M(p) \leq H(p) \sqrt{d+1} $$ as given here, so applying to $g(x)$ we have an inequality $$ H(g) \leq \binom{mn}{\lfloor mn/2\rfloor} M(g) $$ We compute $M(g)$ directly: $$ \begin{align} M(g) &= M(a^nb^m\prod_{j=1}^n\prod_{i=1}^m(x-\alpha_i\beta_j))\\ &=|a|^n|b|^m\prod_{j=1}^n \prod_{i=1}^m \max\{1,|\alpha_i||\beta_j|\}\\ &\leq |a|^n|b|^m\prod_{j=1}^n \prod_{i=1}^m \max\{1,|\alpha_i|\}\max\{1,|\beta_j|\}\\ &=|b|^m\prod_{j=1}^n (\max\{1,|\beta_j|\})^m\cdot |a|\prod_{i=1}^m \max\{1,|\alpha_i|\}\\ &=|b|^m\prod_{j=1}^n (\max\{1,|\beta_j|\})^m M(f_\alpha) \\ &= M(f_\alpha)^n \left(|b|\prod_{j=1}^n \max\{1,|\beta_j|\}\right)^m \\ &= M(f_\alpha)^n M(f_\beta)^m\\ &\leq (\sqrt{m+1}H(f_\alpha))^n (\sqrt{n+1} H(f_\beta))^m \end{align} $$ where the last inequality is from the Mahler measure inequality before. Hence $$ H(g) \leq \binom{mn}{\lfloor mn/2\rfloor} M(g) \leq H(f_{\alpha})^n H(f_{\beta})^m \binom{mn}{\lfloor mn/2 \rfloor} (m+1)^{n/2} (n+1)^{m/2} $$ completing the proof. $$\tag*{$\square$}$$
Finally, putting everything together, $$ \begin{align} H(f_{\alpha\beta}) &\leq H(g) \leq H(f_\alpha)^n H(f_\beta)^m \binom{mn}{\lfloor mn/2\rfloor} (m+1)^{n/2} (n+1)^{m/2}\\ h(f_{\alpha\beta}) &\leq nh(f_\alpha)+mh(f_\beta) + \log \left( \binom{mn}{\lfloor mn/2 \rfloor} (m+1)^{n/2} (n+1)^{m/2}\right)\\ &= nh(f_\alpha) + mh(f_\beta) + C(m,n) \end{align} $$ This gives us the height bound for the algebraic integer $\alpha\beta$ when its degree over $\mathbb Q$ is $mn$.
Remark: The expression $C(m,n)$ seems to be very close to $mn$.
From section 1, we see that minimal polynomial of $f_{\alpha\beta}(x)$ divides $g(x)$ since $g(\alpha\beta)=0$. Hence we may write $$ f_{\alpha\beta}(x) q(x) = g(x) $$ for some $q(x)\in\mathbb Z[x]$. Taking Mahler measure, which is multiplicative, we have $$ M(f_{\alpha\beta}) M(q) = M(g) $$ Since $q(x)$ has integer coefficients, by definition of Mahler measure we see that $$ M(q) \geq 1 $$ Therefore $$ M(f_{\alpha\beta}) \leq M(g) $$ Let $d<mn$ be the degree of $f_{\alpha\beta}(x)$. Applying the Mahler inequality, we have $$ \begin{align} \binom{d}{\lfloor d/2 \rfloor}^{-1} H(f_{\alpha\beta}) &\leq M(f_{\alpha\beta}) \leq M(g) \leq H(g) \sqrt{mn+1}\\ H(f_{\alpha\beta}) &\leq H(g) \binom{d}{\lfloor d/2 \rfloor} \sqrt{mn+1}\leq H(g) \binom{mn}{\lfloor mn/2 \rfloor} \sqrt{mn+1} \end{align} $$ We have shown in section 2 that $$ H(g) \leq H(f_\alpha)^n H(f_\beta)^m \binom{mn}{\lfloor mn/2 \rfloor} (m+1)^{n/2} (n+1)^{m/2} $$ Therefore combining both, we get $$ \begin{align} H(f_{\alpha\beta}) &\leq H(g) \binom{mn}{\lfloor mn/2 \rfloor} \sqrt{mn+1}\\ &\leq H(f_\alpha)^n H(f_\beta)^m \binom{mn}{\lfloor mn/2 \rfloor}^2 \sqrt{mn+1} (m+1)^{n/2} (n+1)^{m/2}\\ h(f_{\alpha\beta}) &\leq nh(f_\alpha) + mh(f_\beta) + \log(\binom{mn}{\lfloor mn/2 \rfloor}^2 \sqrt{mn+1} (m+1)^{n/2} (n+1)^{m/2})\\ &= nh(f_\alpha) + mh(f_\beta) + D(m,n) \end{align} $$ We remark that $D(m,n)$ is "a little" larger than $C(m,n)$, but still a fixed constant once we have chosen the degrees of $f_\alpha(x),f_\beta(x)$.
Picked from choosing small random coefficients $x^3+a_1x+a_0$ and $x^2+b_1x+b_0$. $$ \begin{align} f_\alpha(x) &= x^3+x-17, \alpha\approx 2.44176\\ f_\beta(x) &= x^2+10x-20, \beta = -3+\sqrt 5\\ f_{\alpha\beta}(x) &= x^6+140 x^4+27200 x^3+400 x^2+68000 x-2312000 \end{align} $$ The other two roots of $f_{\alpha}(x)$ are complex.
Now $$ \begin{align} h(f_{\alpha\beta}) &= \log(2312000)\approx 14.6536\\ h(f_\alpha) &= \log(17)\\ h(f_\beta) &= \log(20)\\ nh(f_\alpha) + mh(f_\beta) &\approx 14.6536 \end{align} $$ So $h(f_{\alpha\beta}) \approx nh(f_\alpha) + mh(f_\beta)$.
The "extra" term contributes: $$ C(m,n) = \log(\binom{6}{3} (3+1)^{2/2} (2+1)^{3/2}) \approx 6.02995 $$
Another one, picking random coefficients in $[-2^{16},2^{16}]$: $$ \begin{align} f_{\alpha}(x) &= 26249 + 49279 x + x^3, \alpha \approx -0.532658\\ f_{\beta}(x) &= -55643 + 47743 x + x^2, \beta = \frac{1}{2} \left(-\sqrt{2279616621}-47743\right)\\ f_{\alpha\beta}(x) &= -118701754103013160002707 - 191207722325905327988897 x + 7518736182133771609 x^2 - 2856759510952394846 x^3 + 112331743403465 x^4 + x^6 \end{align} $$ giving heights: $$ \begin{align} h(f_\alpha) &= \log(49279)\\ h(f_\beta) &= \log(55643)\\ nh(f_\alpha)+mh(f_\beta) &\approx 54.3906\\ h(f_{\alpha\beta}) &= \log(191207722325905327988897)\approx 53.6076 \end{align} $$ where $C(m,n)$ is still $\approx 6.02995$.
An example where $h(f_{\alpha\beta}) > nh(f_\alpha) + mh(f_\beta)$. $$ \begin{align} f_\alpha(x) &= -33909+19672 x+x^3\\ f_\beta(x) &= -41717+40527 x+x^2\\ f_{\alpha\beta}(x) &= -83477449753997709876453+47047233759121009092744 x+673477622777766976 x^2+2257259257212603840 x^3+32311676318536 x^4+x^6 \end{align} $$ giving heights $$ \begin{align} h(f_\alpha) &= \log(33909)\\ h(f_\beta) &= \log(41717)\\ h(f_{\alpha\beta}) &= \log(83477449753997709876453)\\ h(f_{\alpha\beta}) - nh(f_\alpha) - mh(f_\beta) &\approx 7.10543*10^{-15} > 0 \end{align} $$
