Exercise 6.3

Question

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If $\alpha$ and $\beta$ are algebraic integers, prove that any solution to $f(x) = x^2 + \alpha x + \beta = 0$ is an algebraic integer. Generalize this result.

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}}

\begin{proof}

Let $\gamma$  be a root of  $x^2 + \alpha x + \beta$, where $\alpha, \beta$ verify :
$$\alpha^n + r_1 \alpha^{n-1}+ \cdots + r_n = 0,\quad r_i \in \mathbb{Z},$$
$$\beta^m + s_1 \beta^{m-1}+ \cdots + s_m= 0,\quad  s_j \in \mathbb{Z}.$$

Let $V$ the set of linear combinations with integer coefficients of 
$$\alpha^i \beta^j \gamma^k, 0\leq i <n,0 \leq j <m,0 \leq k <2.$$

Then $V$ if a finitely generated $\Z$-module.

Moreover, for all  $\delta \in V, \gamma \delta \in V$. Indeed, every $\delta \in V$ is a linear combination with coefficients in $\Z$ of $\alpha^i \beta^j, \alpha^i \beta^j \gamma$, and
 \begin{align*}
 \gamma (\alpha^i \beta^j) &= \alpha^i \beta^j \gamma \in V\
 \gamma(\alpha^i \beta^j \gamma) &= \alpha^i \beta ^j \gamma^2 = \alpha^i \beta^j (-\alpha \gamma - \beta) = - \alpha ^{i+1} \beta^j \gamma - \alpha^i \beta^{j+1} \in V.
 \end{align*}
 (if $i+1 = n$, we replace $\alpha^{i+1} = \alpha^n$ by $ -\sum\limits_{k=1}^{n-1} r_k \alpha^{n-k}$, and a similar replacement if if $j+1 = m$.)
 
 As for each $x\in V$, where $V$ if a finitely generated $\Z$-module, $x \gamma \in V$, so  $\gamma$ is an algebraic integer (Proposition 6.1.4).

More generally,  if  $\gamma^n + \alpha_1 \gamma^{n-1}+ \cdots + \alpha_n = 0$, where the $\alpha_i$ are algebraic integers, then  $x$ is an algebraic integer.
\end{proof}

Exercise 6.3

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Exercise 6.3

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