most recent 30 from http://mathoverflow.net 2013-05-26T04:40:18Z http://mathoverflow.net/feeds/user/16864 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71784/comparing-the-height-of-an-algebraic-number-with-the-height-of-its-conjugates Dali 2011-08-01T07:53:05Z 2011-08-01T15:28:06Z

<p>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 <code>$$H(B) := \max \{H(\alpha) \ | \ \alpha\in B\}.$$</code></p> <p>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 $\alpha$.</p> <p><strong>Question.</strong> Can we bound $H(B)$ from above in terms of data depending only on $\alpha$ and $K$? </p> <p><strong>Example.</strong> The number of elements of $B$ is less or equal to $[K:\mathbf{Q}]$.</p> <p>I'm looking for a bound of the form $H(B)\leq H(\alpha)^{[K:\mathbf{Q}]}$ if possible.</p>

http://mathoverflow.net/questions/71784/comparing-the-height-of-an-algebraic-number-with-the-height-of-its-conjugates/71792#71792 Dali 2011-08-01T10:44:59Z 2011-08-01T10:44:59Z

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.