Exercise 6.6

Proof. By definition, for all $z\in ℂ,z\in \mathbb{Q}[\alpha ]⟺\exists <!--\; FUNCTION\; APPLICATION\; -->P\in \mathbb{Q}[x],z=P(\alpha )$.

The Euclidean division gives $P=Q_1(x^2+\mathit{mx}+n)+R,Q_1,R\in \mathbb{Q}[x],\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(R)<2$, so $R=\mathit{rx}+s,r,s\in \mathbb{Q}$. Therefore $z=Q_1(\alpha )({\alpha }^2+\mathit{m\alpha }+n)+\mathit{r\alpha }+s=\mathit{r\alpha }+s$ :

$\text{∙}$ $\mathbb{Q}[\alpha ]\subset ℂ$, where $(ℂ,+,\times )$ is a field. $1\in \mathbb{Q}[\alpha ]$ ( $1=P_0(\alpha )$, where $P_0$ is the constant polynomial $1$).

$\text{∙}$ Let $\beta ,\gamma \in \mathbb{Q}[\alpha ]:\beta =P(\alpha ),\gamma =Q(\alpha )$, where $P,Q$ are in $\mathbb{Q}[x]$. Then $\alpha -\beta =P(\alpha )-Q(\alpha )=R(\alpha )$, where $R=P-Q\in \mathbb{Q}[x]$, and $\mathit{\alpha \beta }=P(\alpha )Q(\alpha )=S(\alpha )$, where $S=\mathit{PQ}\in \mathbb{Q}[x]$. Thus $\alpha -\beta \in \mathbb{Q}[\alpha ],\mathit{\alpha \beta }\in \mathbb{Q}[\alpha ]$. So $\mathbb{Q}[\alpha ]$ is a subring of $(ℂ,+,\times )$.

$\text{∙}$ Let $\beta =P(\alpha )\in \mathbb{Q}[\alpha ],P\in \mathbb{Q}[x]$ and $\beta \ne 0$. As $\beta \ne 0$, $Q=x^2+\mathit{mx}+n\nmid P$.

Let $D\in \mathbb{Q}[x]$ such that $D\mid P,D\mid Q$. As $Q$ is irreducible by hypothesis, $D=\lambda$ or $D=\mathit{\lambda Q},\lambda \in {ℂ}^{\text{∗}}$ ( $D$ is an associate of $1$ or $Q$). If $D=\mathit{\lambda Q}$, then $Q\mid D$, and $D\mid P$, so $Q\mid P$. Since $Q(\alpha )=0$, this implies $\beta =P(\alpha )=0$, in contradiction with the definition of $\beta$. So $D=\lambda \mid 1$. Therefore $P\wedge Q=1$.

From Bézout’s theorem, there exist polynomials $U,V\in \mathbb{Q}[x]$ such that $\mathit{UP}+VQ=1$. As $\mathbb{Q}(\alpha )=0$, $U(\alpha )P(\alpha )=1$ and $\gamma =U(\alpha )\in \mathbb{Q}[\alpha ]$ is such that $\mathit{\gamma \beta }$ = 1. Therefore $\mathbb{Q}[\alpha ]$ is a subfield of $(ℂ,+,\times )$ (and $\mathbb{Q}(\alpha )=\mathbb{Q}[\alpha ]$).

As $x^2+\mathit{mx}+n$ is irreducible, $\mathrm{\Delta }=m^2-4n\ne 0$ (if not, $x^2+\mathit{mx}+n={(x+m∕2)}^2-(m^2-4n)∕4={(x+m∕2)}^2$ is not irreducible). So $\mathrm{\Delta }\in \mathbb{Z}\setminus \{0\}$ can be written $\mathrm{\Delta }=m^2-4n=D_0^{2}D$, where $D$ is square-free (positive or negative), $D\ne 0,D_0\ne 0$.

$\alpha =-\frac{m}{2}+𝜀\frac{\sqrt{\mathrm{\Delta }}}{2},𝜀=\pm 1$, so $\alpha =-\frac{m}{2}+𝜀D_0\frac{\sqrt{D}}{2}$, thus $\alpha \in \mathbb{Q}[\sqrt{D}]$ and $\mathbb{Q}[\alpha ]\subset \mathbb{Q}[\sqrt{D}]$.

As $D_0\ne 0$, $\sqrt{D}=𝜀\frac{2\alpha +m}{D_0}\in \mathbb{Q}[\alpha ]$, so $\mathbb{Q}[\sqrt{D}]\subset \mathbb{Q}[\alpha )]$ :

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