Exercise 6.2

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $\alpha$ be an algebraic number. Show that there's an integer $n$ such that $n\alpha$ is an algebraic integer.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

($0$ is a valid answer to this sentence ! More seriously, we search a {\it positive} integer $n$.)

\begin{proof}

Let  $\alpha$ an algebraic number.  By definition, there exist $a_0,a_1,\cdots,a_n \in \Z, a_n 
eq 0,$ such that
$$a_n \alpha^n+ a_{n-1} \alpha^{n-1} + \cdots + a_k \alpha^k+\cdots +a_0 = 0.$$
(Up to multiply this equation by $-1$, we can suppose that $a_n >0$).

Multiplying by $a_n^{n-1}$, we obtain
$$a_n^n \alpha^n+ a_n^{n-1}a_{n-1} \alpha^{n-1} + \cdots + a_n^{n-1} a_k \alpha^k+\cdots + a_n^{n-1} a_0 = 0.$$
So
$$(a_n \alpha)^n+ a_{n-1} (a_n\alpha)^{n-1} + \cdots + a_n^{n-k-1}a_k (a_n \alpha)^k+\cdots + a_n^{n-1} a_0 = 0.$$

Soit $p(x) = x^n +\sum\limits_{k=0}^{n-1} a_n^{n-k-1}a_k x^k$. Then $p(x) \in \mathbb{Z}[x]$, $p(x)$ is monic, and $p(a_n \alpha) = 0$. So $a_n \alpha$ is an algebraic integer, with $m = a_n \in \mathbb{N}^*$.

Conclusion : if  $\alpha$ is an algebraic number, there exists an integer $m >0$ such that $m\alpha$ is an algebraic integer.
\end{proof}

Exercise 6.2

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Exercise 6.2

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