The definition of Weil height (or absolute and logarithmic height) that I am familiar with is the following:
For an algebraic integer $\alpha$ and a finite extension $K$ of $\mathbb Q$ containing $\alpha$, we define the logarithmic height $h(\alpha)$ of $\alpha$ by
$$h(\alpha) := \sum_{v \in M_K} \frac{[K_v:\mathbb Q_v]}{[K:\mathbb Q]} \log \max \{1, |\alpha|_v \}$$
where $M_K$ is a set of representatives of equivalence classes of absolute values on $K$ satisfying the product formula and $K_v$ (respectively $\mathbb Q_v$) denotes the completion of $K$ (respectively $\mathbb Q$) with respect to the place $v \in M_K$.
I want to show that for any algebraic number $\alpha$ which is not an algebraic integer $$h(\alpha) \geqslant \frac{\log 2}{[\mathbb Q (\alpha) :\mathbb Q]} \hspace{5mm} \cdots (1)$$
I could show, by embedding $K$ in its Galois closure over $\mathbb Q$, that $h(\alpha) = h(\beta)$ for all other roots $\beta$ of the minimal polynomial (say $f(x) \in \mathbb Q[x]$) of $\alpha$ over the rationals. So, we would be done if every algebraic number $\alpha$ (which is not an algebraic integer) has a conjugate (i.e. some other root of its minimal polynomial over $\mathbb Q$) $\beta$ satisfying $|\beta| \geq 2$ (where $|\cdot|$ here denotes the usual archimedean absolute value on $\mathbb C$). But I'm not sure if such a $\beta$ must always exist for algebraic non-integers $\alpha$ (as an aside, does it?) and perhaps there is a simpler and more direct way of showing (1). I would appreciate any help.
Best Answer
If $\alpha$ is not integral, there is some finite place $v$ with $v(\alpha) \leq -1$ and hence a prime $p$ such that $|\alpha|_v \geq p\geq 2$. Also it is clear that $[K_v:\mathbb Q_v] \geq 1$
Take $K=\mathbb Q$ in your definition of the height, and look at just this place: $$\frac{[K_v:\mathbb Q_v]}{[\mathbb Q(\alpha):\mathbb Q]}\log\max\{1,|\alpha|_v\} \geq\frac{\log 2}{[\mathbb Q(\alpha):\mathbb Q]}$$
The rest of the terms in the sum are all nonnegative and so the inequality follows.