# The house of an algebraic number

Let $$\alpha$$ be a non-zero algebraic number of degree $$d$$. Denote $${\rm den}(\alpha)$$ the smallest positive integer $$m$$ such that $$m\alpha$$ is an algebraic integer, and $${\rm House}(\alpha)$$ the maximal absolute value among the conjugates of $$\alpha$$. Show that $$\vert \alpha \vert \geq {\rm den}(\alpha)^{-d}\cdot{\rm House}(\alpha)^{1-d}$$.

I have tried several approach but they led to nothing. For instance let $$F(X)=a_0X^d+a_1X^{d-1}+...+a_{d-1}X+a_d$$ be the minimal primitive polynomial in $$\mathbb{Z}[X]$$ that receive $$\alpha$$ as a root, and let $$\alpha_1,...,\alpha_d$$ be the conjugates of $$\alpha$$. Then $${\rm den}(\alpha)^{d}\cdot{\rm House}(\alpha)^{d-1}\cdot\vert \alpha \vert \geq {\rm den}(\alpha)^{d}\cdot \prod_{i=1}^d \vert \alpha_i \vert \geq {\rm den}(\alpha)^{d}\cdot \left\vert\frac{a_n}{a_0}\right\vert,$$ but then I don't know what to do next. Another approach is that considering two case $$\vert \alpha \vert < 1$$ and $$\vert \alpha \vert \geq 1$$. Again the second case is easy but I have no idea to solve the first one. It is like the directions of the inequalities conflict and I don't know any evalution that would overcome this issue.

Any help is appreciated.

• Can you define the house of $\alpha$? Dec 27, 2023 at 14:03
• I edited with a brief defintion of the house. Thanks for pointing out :D Dec 27, 2023 at 14:21

By multiplying the inequality by $$\operatorname{den}(\alpha)$$ we may assume $$\operatorname{den}(\alpha)=1$$, i.e. $$\alpha$$ is an algebraic integer. So we have to show $$|\alpha|\geq\operatorname{House}(\alpha)^{1-d}$$. Let $$b_1,\dots,b_d$$ be the absolute values of the conjugates of $$\alpha$$ and wlog $$b_1= |\alpha|$$. Let $$i$$ be such that $$b_i=\operatorname{House}(\alpha)=\max_{j=1,\dots,d}b_j$$. Then $$b_i^{d-1}b_1\geq b_1b_2\cdots b_d=|N_{\Bbb Q(\alpha)/\Bbb Q}(\alpha)|\geq1,$$ so $$|\alpha|\geq\operatorname{House}(\alpha)^{1-d}$$.

• Just a small question: why can we assume that the ${\rm den}(\alpha)=1$? Dec 28, 2023 at 7:41
• Let $m=\operatorname{den}(\alpha)$. Then the inequality in question is equivalent to $|m\alpha|\geq m\operatorname{den}(\alpha)^d\operatorname{House}(\alpha)^{1-d}=m^{1-d}\operatorname{House}(\alpha)^{1-d}=\operatorname{House}(m\alpha)^{1-d}$ (the conjugates of $m\alpha$ are $m$ times the conjugates of $\alpha$). This is just the desired inequality for $m\alpha$ instead of $\alpha$ (as $\operatorname{den}(m\alpha)=1$). Dec 28, 2023 at 15:16
• Ah yeh I also just realized the notes about conjugates of $m\alpha$. Thanks :D Dec 28, 2023 at 15:20