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I would like to find effective upper bound for the height of $a+b$ and $a/b$ and $ab$ knowing the heights of $a$ and $b$. Thanks.

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Sorry, what is the height? – Martin Brandenburg May 11 2011 at 18:59

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If you know only heights of $a$ and $b$, you may estimate heights of $a+b$, $a/b$ and $ab$. Assuming that $h$ is an absolute (Weil) height: $$h(ab)\leq h(a)+h(b)$$ $$h(a/b)\leq h(a)+h(b)$$ $$h(a+b)\leq\log 2 +h(a)+h(b)$$ This bounds are sharp. You may find this, for example, in M. Waldschmidt "Diophantine approximation on linear algebraic groups", Chapter 3.

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$h(a)$ is also called the absolute logarithmic height of $a$, defined by $1/[K:Q] \log H_K(a)$, where $H_K(a)$ is the product of all of the embeddings (with multiplicity) of $a$ that land outside the complex unit circle. In other words, $h(a) = 1/d M(a)$, where $M(a)$ is the Mahler measure of $a$. – Kevin O'Bryant May 29 at 13:55
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There are standard estimates for the heights of algebraic numbers $a_1,...,a_n$ in terms of their elementary symmetric functions $s_1,...,s_n$, or equivalently, estimates relating the heights of the roots of a polynomial to the heights of its coefficients. You can find this in many textbooks, including for example my Arithmetic of Elliptic Curves (Theorem VIII.5.9) or Lang's Fundamentals of Diophantine Geometry (Chapter 3, Section 2). The estimate is $$ \sum_{i=1}^n h(a_i) - n\log(2) \le h([1,s_1,...,s_n]) \le \sum_{i=1}^n h(a_i) + (n-1)\log(2). $$

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If $a$ is the root of a characteristic polynomial of $M$ and $b$ is the root of the characteristic polynomial of $N,$ then $ab$ is a root of the charcteristic polynomial of $M \otimes N,$ and $a+b$ is a root of the characteristic polynomial of $M \otimes I + I \otimes N.$ That should be enough to compute the height.

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Thanks, Igor. Can you give the definition of the tensor product of two polynomials. – vanvu May 11 2011 at 17:22
M and N are matrices, and the tensor product is the thing they call the block product. – James Cranch May 11 2011 at 17:32
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The tensor product is of MATRICES $M$ and $N,$ of which $a$ and $b$ are eigenvalues (you can take $M$ and $N$ to be the companion matrices of the minimal polynomials of $a, b$). – Igor Rivin May 11 2011 at 17:32
Sorry to duplicate, Igor! – James Cranch May 11 2011 at 17:33
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No problem, I was going to apologize to you for same... – Igor Rivin May 11 2011 at 17:34
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I too was looking for the answer to the same question. It seems necessary to define "height" since there are of course several variants in use. The height $h(a)$ I am interested in is the maximum of the absolute values of the coefficients of the minimal polynomial of the algebraic number $a$. Note that the first answer refers to a different notion of height, since, for example,

$9 = h(9) = h(3 \cdot 3) \not \leq h(3) + h(3) = 3 + 3 = 6.$

I imagine there must be upper bounds of the form

$h(ab) \leq f(d) h(a)^{g(d)} h(b)^{g(d)}$

for some simple functions $f$ and $g$, where $d$ is the degree of a field extension of $\mathbb{Q}$ containing both $a$ and $b$, for example.

Are there any such results in the literature, and similarly for $h(a+b)$ and $h(a/b)$?

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I think Oleg's height is the log of your height. – Will Sawin May 28 at 21:05
Thanks Will. I believe Oleg's height is the logarithmic Weil height, defined in terms of the Mahler measure. It is no doubt related to my "naive" height in some way, but I don't think it's simply the log of it. Or at least, it's far from obvious to me... – Joel Ouaknine May 28 at 21:32
This height is the "usual height", and is often written with a capital $H$ to distinguish it from the absolute logarithmic height. Waldschmidt (Lemma 3.11) reports $\frac1d H(a) - \log 2 \leq h(a) \leq \frac1d H(a) + \frac{1}{2d} \log(d+1)$, which can be combined with the bounds on other answers to bound $H(a+b)$ and $H(a/b)$. – Kevin O'Bryant May 29 at 14:07
Thanks very much Kevin for the pointer. I just took Waldschmidt's book out and indeed it contains all the results I need! (By the way, your statement of Lemma 3.11 is missing an application of $\log(-)$ to $H(a)$ on both instances.) – Joel Ouaknine May 30 at 0:52

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