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Let $\bar{\mathbf{Q}}$ be an algebraic closure of the rationals, and $\alpha$ denote an algebraic number in $\bar{\mathbf{Q}}$. We define the height of $\alpha$, denoted by $H(\alpha)$, to be $$H(\alpha) = \left( \prod_v \max(1,\Vert \alpha\Vert_v) \right)^{1/[K:\mathbf{Q}]}.$$ Here $K$ is a number field containing $\alpha$ and the product runs over the set of normalized valuations $v$ of $K$. For a non-empty finite subset $B\subset \bar{\mathbf{Q}}$, we define $$H(B) := \max \{H(\alpha) \ | \ \alpha\in B\}.$$

Now, let $K$ be a number field and let $\alpha$ be an algebraic number contained in $K$. Let $B$ be the set of conjugates of $\alpha$.

Question. Can we bound $H(B)$ from above in terms of data depending only on $\alpha$ and $K$?

Example. The number of elements of $B$ is less or equal to $[K:\mathbf{Q}]$.

I'm looking for a bound of the form $H(B)\leq H(\alpha)^{[K:\mathbf{Q}]}$ if possible.

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up vote 7 down vote accepted

One always has $H(B)=H(\alpha)$. Indeed, since presumably you normalise the absolute values in such a way that the definition does not depend on the choice of $K$ for a given $\alpha$ (as long as $\alpha\in K$), let's replace $K$ by its Galois closure $K'$ over $\mathbb{Q}$. Let $\sigma$ be any element in the Galois group. Since $\sigma$ acts by permutation on the set of absolute values of $K'$, $H(\sigma(\alpha))$ is just the same product as $H(\alpha)$ with some of the terms permuted.

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Nice. I didn't realize the elements of the Galois group permute the set of absolute values of the Galois closure of $K$. Thanks alot. – Dali Aug 1 '11 at 10:44

Alex has answered your question, but if you're working with heights and Galois conjugates of numbers, you're likely at some point to need a relation between the height of $\alpha$ and the heights of the elementary symmetric functions $s_1,\ldots,s_n$ of its conjugates $\alpha_1,\ldots,\alpha_n$. The advantage of working with $s_1,\ldots,s_n$, of course, is that they are in $\mathbb{Q}$. A useful relation (the constants can be improved) is $$\frac{1}{2}H(\alpha) \le H([1,s_1,\ldots,s_n])^{1/n} \le 2H(\alpha).\qquad(*)$$ This is proven in many standard texts, but for concreteness I can point to my book The Arithmetic of Elliptic Curves, Theorem VIII.5.9. (To get $(*)$ from that theorem, you also need to use the fact, as noted in Alex's answer or see [AEC, Theorem VIII.5.10], that Galois conjugates have the same height.)

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