Exercise 6.5

Proof. As $\alpha$ is an algebraic integer, there exists a monic polynomial $h(x)\in \mathbb{Z}[x]$ such that $h(\alpha )=0$. As $f(x)\in \mathbb{Q}[x]$ is the minimal polynomial of $\alpha$, and $h(\alpha )=0$, $f(x)$ divides $h(x)$ in $\mathbb{Q}[x]$.

(Quick reminder : $h(x)=q(x)f(x)+r(x),q(x),r(x)\in \mathbb{Q}[x],\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(r(x))<\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f(x))$ or $r(x)=0$. As $r(\alpha )=0$ and $f(x)\in \mathbb{Q}[x]$ is the monic polynomial of least degree such that $f(\alpha )=0$, $r=0$ so $f(x)\mid h(x)$).

So there exists $g(x)\in \mathbb{Q}[x]$ such that $h(x)=f(x)g(x)$. As $h(x),f(x)$ are both monic, $g(x)$ is also monic.

Let $d\in \mathbb{Z},d\ne 0$ such that $\mathit{df}(x)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^m{a}_i{x}^i\in \mathbb{Z}[x]$, and $c=a_1\wedge a_2\wedge \cdots \wedge a_m$, $a_i=c{b}_i$, with $b_1\wedge b_2\wedge \cdots \wedge b_m=1$, so $f(x)=\frac{c}{d}{f}_1(x)$, where $f_1$ is primitive. Similarly $g(x)=\frac{s}{t}{g}_1(x)$, $s,t\in \mathbb{Z}$, $g_1(x)$ primitive.

So $h(x)=\frac{\mathit{cs}}{\mathit{dt}}{f}_1(x)f_2(x)=\frac{u}{v}{f}_1(x)f_2(x)$, where $u\wedge v=1,v>0$. The polynomial $f_1(x)f_2(x)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{k=0}^r{c}_k{x}^k$ is primitive (Ex. 6.4). As $\mathit{vh}(x)=u{f}_1(x)f_2(x)$, $v\mid u{c}_k$ , and $u\wedge v=1$, thus $v\mid c_k,k=0,1,\dots ,r$. As $c_1\wedge \cdots c_k=1$, $v\mid 1$, where $v>0$, so $v=1$. $h(x)=u{f}_1(x)f_2(x)$ is monic, thus $u=\pm 1$, and $\pm f_1,\pm f_2$ are monic. From $f(x)=\frac{c}{d}{f}_1(x)$ we deduce $\frac{c}{d}=\pm 1$ and $f(x)=\pm f_1(x)\in \mathbb{Z}[x]$.

Conclusion : if $f(x)$ is the minimal polynomial of an algebraic integer $\alpha$, $f\in \mathbb{Z}[x]$. □