# Proof of remark in algebraic number theory

$$\def\aroof{\overline{\hspace{-1pt}\smash[t]{|}\hspace{1pt}\alpha\hspace{1pt}\smash[t]{|}\hspace{-1pt}}}$$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?

• Please use a high-level tag like "nt.number-theory". I added this tag now, and removed the tag "transcendental-number-theory" as it was not appropriate. Dec 8, 2023 at 21:07
• I know that it's fashionable to refer to results as "remarks" but I think it's a bit confusing. Words are not missing (theorem, proposition, lemma, fact, claim...).
– YCor
Dec 9, 2023 at 8:16

$$\def\aroof{\overline{\hspace{-1pt}\smash[t]{|}\hspace{1pt}\alpha\hspace{1pt}\smash[t]{|}\hspace{-1pt}}}$$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 $$\left|a_i\right| \le C\sum_{j=1}^h\left|\alpha^{(j)}\right|\le C\,h\;\aroof.$$
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.