Proof of remark in algebraic number theory

Question

$\def\aroof{\overline{\hspace{-1pt}\smash[t]{|}\hspace{1pt}\alpha\hspace{1pt}\smash[t]{|}\hspace{-1pt}}}$In Hua Loo Keng's "Introduction to Number Theory" page 489, there is the following remark:

Let $K$ be an algebraic number field of degree $h,$ and let $\beta_1, \ldots , \beta_h$ be an integer basis, so that every integer in $K$ has the unique representation $a_1\beta_1 + \ldots + a_h\beta_h$ where $a_1, \ldots , a_h$ are rational integers.

We shall denote by $\aroof$ the maximum of the modulus of the conjugates $\alpha^{(i)}$ with $(1 \leq i \leq h)$ of $\alpha,$ that is $$\aroof = \max_{1 \leq i \leq h} \left|\alpha^{(i)}\right|.$$

In the following we let $c$ be a natural number depending on $K$ and its basis $\beta_1, \ldots , \beta_h.$

It can be shown that if $\alpha$ is an algebraic integer with $\alpha = a_1\beta_1 + \ldots + a_h\beta_h,$ then $$\left|a _i\right| \leq c \;\aroof·$$ What is the proof of this statement?

Answer 1

$\def\aroof{\overline{\hspace{-1pt}\smash[t]{|}\hspace{1pt}\alpha\hspace{1pt}\smash[t]{|}\hspace{-1pt}}}$One has the $h\times h$ linear system
$$\alpha^{(i)} = a_1\beta_1^{(i)} + \ldots + a_h\beta_h^{(i)},\quad 1\le i\le h.$$ Let $B$ be the $h\times h$ matrix with coefficients $B_{ij}=\beta_j^{(i)}$. Then $B$ is invertible (the square of its determinant is the discriminant of $K$). Since $(\alpha^{(1)},\dots,\alpha^{(n)})^{T}=B(a_1,\dots,a_h)^T$ one obtains that $$(a_1,\dots,a_h)^T=B^{-1}(\alpha^{(1)},\dots,\alpha^{(n)})^{T}.$$ Let $c_{ij}$ be the coefficients of $B^{-1}$ which depend only on the basis $\beta_1,\dots,\beta_h$. Then $$\left|a_i\right|=\left|\sum_{j=1}^hc_{ij}\alpha^{(j)}\right| \le\sum_{j=1}^h\left|c_{ij}\alpha^{(j)}\right|.$$ So if we set $C=\max_{ij}\left|c_{ij}\right|$ then $$\left|a_i\right| \le C\sum_{j=1}^h\left|\alpha^{(j)}\right|\le C\,h\;\aroof.$$

Answer 2

Consider the distinct embeddings $\sigma_\ell\colon K \to \mathbb{C}$ of the number field $K$ in the complex numbers (up to equality on $K$, not just up to their image): by standard facts in Galois theory, there are exactly $h$ (i.e. $[K:\mathbb{Q}] =: h$) of them, and they are linearly independent over $\mathbb{C}$ (see: linear independence of characters). Now each $\sigma_\ell$ is a $\mathbb{Q}$-linear map $a_1\beta_1 + \cdots + a_h\beta_h \mapsto b_{\ell,1}\, a_1 + \cdots + b_{\ell,h}\, a_h$ for some complex coefficients $b_{\ell,j} := \sigma_\ell(\beta_j)$: linear independence tells us that the $h\times h$ complex matrix $(b_{\ell,j})$ is invertible, say $\sum_{\ell=1}^h c_{i,\ell}\, b_{\ell,j} = \delta_{i,j}$, and then $a_i = \sum_{\ell,j} c_{i,\ell}\, b_{\ell,j}\, a_j$ for any complex $a_1,\ldots,a_h$ and in particular for rational ones, meaning $a_i = \sum_\ell c_{i,\ell} \, \sigma_\ell(\alpha)$ for $\alpha\in K$, which gives us $|a_i| \leq c\cdot \max|\sigma_\ell(\alpha)|$ where $c = \lceil\sum_\ell |c_{i,\ell}|\rceil$, as claimed.