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.



Comparing the height of an algebraic number with the height of its conjugates
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.


