Exercise 7.3.10

Proof.

(a) $\Rightarrow$(b):

If $\mathbb{Q}(\alpha )=\mathbb{Q}(\beta )$, then $\alpha \in \mathbb{Q}(\beta )$. Since $[\mathbb{Q}(\beta ):\mathbb{Q}]=2$, $\beta \notin \mathbb{Q}={\mathrm{Vect}}_{\mathbb{Q}}(1)$, then $(1,\beta )$ is a linearly independent list with 2 elements in a 2-dimensional vector space, so is a basis of $\mathbb{Q}(\beta )$ over $\mathbb{Q}$. Then $\alpha$ spans on this basis under the form

Moreover, $b\ne 0$, otherwise $\alpha \in \mathbb{Q}$, and $\alpha$ would not be of degree 2 over $\mathbb{Q}$.

(b) $\Rightarrow$(a):

If $\alpha =a+\mathit{b\beta },a,b\in \mathbb{Q},b\ne 0$, then $\alpha \in \mathbb{Q}(\beta )$, thus $\mathbb{Q}(\alpha )\subset \mathbb{Q}(\beta )$.

Moreover $\beta =b^{-1}(\alpha -a)\in \mathbb{Q}(\alpha )$, so $\mathbb{Q}(\beta )\subset \mathbb{Q}(\alpha )$.

(b) $\Rightarrow$(c):

$\delta =\alpha +\beta =a+(b+1)\beta \in \mathbb{Q}(\beta )$. Therefore the list $(1,\delta ,{\delta }^2)$ of 3 vectors in a 2-dimensional vector space is linearly dependent over $\mathbb{Q}$, so there exist $(u,v,w)\in {\mathbb{Q}}^3\setminus \{(0,0,0)\}$ such that $u{\delta }^2+\mathit{v\delta }+w=0$.

Let $f(x)=u{x}^2+\mathit{vx}+w\in \mathbb{Q}[x]$. Then $f(\alpha +\beta )=0$, with $f\ne 0,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)\le 2$. If $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=2$, (c) is proved.

But $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)<2$ is a possibility, for instance if $\beta =-\alpha$. As $f\ne 0$, then $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=0$ is in contradiction with $f(\delta )=0$, so in this case $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=1$: $f(x)=\mathit{vx}+w,v\ne 0$. Then $\delta =\alpha +\beta$ is a root of the polynomial of degree 2 $x(\mathit{vx}+w)$. In both cases,

$\alpha +\beta$ is the root of a quadratic polynomial in $\mathbb{Q}[x]$.

(c) $\Rightarrow$(a): Suppose that $\alpha +\beta$ is a root of a quadratic polynomial, and suppose at the contrary that $\mathbb{Q}(\alpha )\ne \mathbb{Q}(\beta )$. By assumption, $[\mathbb{Q}(\alpha ):\mathbb{Q}]=[\mathbb{Q}(\beta ):\mathbb{Q}]=2$. Therefore $[\mathbb{Q}(\alpha ,\beta ):\mathbb{Q}(\beta )]\le 2$. If $[\mathbb{Q}(\alpha ,\beta ):\mathbb{Q}(\beta )]=1$, then $\alpha \in \mathbb{Q}(\beta )$, so $\alpha =a+\mathit{b\beta }$ for some $a,b\in \mathbb{Q}$, and $b\ne 0$ otherwise $\alpha \in \mathbb{Q}$.

The implication (b) $\Rightarrow$ (a) shows that $\mathbb{Q}(\alpha )=\mathbb{Q}(\beta )$, and this is a contradiction. Therefore $[\mathbb{Q}(\alpha ,\beta ):\mathbb{Q}(\beta )]=2$, and

Write $L=\mathbb{Q}(\alpha ,\beta )$. Then $[L:\mathbb{Q}]=4$.

Let $f=x^2+\mathit{rx}+s,g=x^2+r^{\prime }x+s^{\prime }\in \mathbb{Q}[x]$ be the minimal polynomials of $\alpha ,\beta$ over $\mathbb{Q}$, and write $\alpha ,{\alpha }^{\prime }$ the roots of $f$, $\beta ,{\beta }^{\prime }$ the root of $g$.

As $\alpha +{\alpha }^{\prime }=-r\in \mathbb{Q}$, ${\alpha }^{\prime }\in \mathbb{Q}(\alpha )$, and similarly ${\beta }^{\prime }\in \mathbb{Q}(\beta )$. Therefore the splitting field of $\mathit{fg}$ is $\mathbb{Q}(\alpha ,{\alpha }^{\prime },\beta ,{\beta }^{\prime })=\mathbb{Q}(\alpha ,\beta )$. This shows that $\mathbb{Q}\subset \mathbb{Q}(\alpha ,\beta )$ is a normal extension, and also separable since the characteristic of $\mathbb{Q}$ is 0. So $\mathbb{Q}\subset L$ is a Galois extension, therefore

Consequently, if we write $G=\mathrm{Gal}(L∕\mathbb{Q})$,

$\text{∙}$ If $G\simeq \mathbb{Z}∕4\mathbb{Z}$, as $\mathbb{Z}∕4\mathbb{Z}$ has a unique subgroup $H$ of index 2 in $G$, there exists a unique quadratic extension of $\mathbb{Q}$ included in $L$, and so $\mathbb{Q}(\alpha )=\mathbb{Q}(\beta )=L_H$, in contradiction with the hypothesis.

$\text{∙}$ We suppose now that $G\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$. Then by the Galois correspondence, the extension $\mathbb{Q}(\alpha )$ corresponds to a subgroup $H$ of index 2 in $G$, thus of order $4∕2=$ 2. So $H=\{e,\sigma \}$, and $\mathbb{Q}(\alpha )=L_H$ is the fixed field of $\sigma$. Similarly there exists $\tau \in G,\tau \ne \sigma$, such that $\mathbb{Q}(\beta )$ is the fixed field of $K=\{e,\tau \}$. There exist exactly 3 subgroups of $G$ of index 2 : $⟨\sigma ⟩,⟨\tau ⟩,⟨\zeta ⟩$, where $\zeta =\mathit{\sigma \tau }=\mathit{\tau \sigma }$, in correspondence with 3 quadratic sub-extensions of $\mathbb{Q}\subset L$, two of them being $\mathbb{Q}(\alpha ),\mathbb{Q}(\beta )$. As every quadratic extension of $\mathbb{Q}$, the third is of the form $\mathbb{Q}(\gamma ),\gamma \in L$, fixed field of $\{e,\zeta \}$.

We show that $\mathbb{Q}(\alpha +\beta )=\mathbb{Q}(\gamma )$. We know that $\alpha +\beta \notin \mathbb{Q}$, otherwise $\mathbb{Q}(\alpha )=\mathbb{Q}(\beta )$. Hence $\mathbb{Q}(\alpha +\beta )$ is a quadratic extension of $\mathbb{Q}$, therefore is equal to $\mathbb{Q}(\alpha ),\mathbb{Q}(\beta )$ or $\mathbb{Q}(\gamma )$.

$\mathbb{Q}(\alpha +\beta )\ne \mathbb{Q}(\beta )$, otherwise $\alpha \in \mathbb{Q}(\beta )$, and so $\mathbb{Q}(\alpha )\subset \mathbb{Q}(\beta )$, where $[\mathbb{Q}(\alpha ):\mathbb{Q}]=[\mathbb{Q}(\beta ):\mathbb{Q}]=2$, thus $\mathbb{Q}(\alpha )=\mathbb{Q}(\beta )$.

Similarly $\mathbb{Q}(\alpha +\beta )\ne \mathbb{Q}(\alpha )$. It remains only the possibility $\mathbb{Q}(\alpha +\beta )=\mathbb{Q}(\gamma )$, fixed field of $\zeta =\sigma \circ \tau$.

Note that $\tau (\alpha )\ne \alpha$, otherwise $\alpha \in L_{⟨\tau ⟩}=\mathbb{Q}(\beta )$, which is excluded.

As $\alpha +\beta \in \mathbb{Q}(\gamma )=L_{⟨\mathit{\sigma \tau }⟩}$,

But, since $G$ is commutative, we have also

Therefore $\alpha +\beta =\tau (\alpha )+\sigma (\beta )$, thus $\alpha -\tau (\alpha )=\sigma (\beta )-\beta$.

As $\mathbb{Q}(\alpha )$ is a normal extension, $\tau (\alpha )\in \mathbb{Q}(\alpha )$, and similarly $\sigma (\beta )\in \mathbb{Q}(\beta )$. Therefore

Thus $\sigma (\beta )=\beta +c,\tau (\alpha )=\alpha -c,c\in {\mathbb{Q}}^{\text{∗}}$.

Then $(\mathit{\sigma \tau })(\alpha +\beta )=\alpha +\beta$, so the orbit of $\alpha +\beta$ under the action of $G=\{e,\sigma ,\tau ,\mathit{\sigma \tau }\}$ is ${\mathcal{}O}_{\alpha +\beta }=\{\alpha +\beta ,\alpha +\beta +c,\alpha +\beta -c\}$ has exactly 3 elements. As the cardinality of the orbit is the index of the stabilizer of $\alpha +\beta$ in $G$, so divides the order of $G$, we would have $3\mid 4=|G|$: this is a contradiction, obtained under the hypothesis $\mathbb{Q}(\alpha )\ne \mathbb{Q}(\beta )$, so

(c) $\Rightarrow$(a) is proved. □