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If $\alpha, \beta$ are algebraic over $\mathbb{Q}$ of degree $2$ and $\alpha +\beta$ is a root of a quadratic polynomial over $\mathbb{Q}$, then prove that $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.

I'm stuck at showing the above statement. I think we have to show that $\alpha$ can be written as $p+q\beta$ or vice versa. But I just can't seem to show that.

I tried something as shown below but got nowhere:

If $g(x)$ is the polynomial of $\alpha+\beta$, $g(x)=f(x)Q(x)+R(x)$ where $f(x)$ is the polynomial of $\beta$

Ideas Anyone?

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Let $\alpha, \beta \in \mathbb{C}$ be algebraic with degree $2$ such that $\alpha +\beta$ is also algebraic of degree $2$. Suppose that $\mathbb{Q}(\alpha) \neq \mathbb{Q}(\beta)$. Then we have the following non-trivial stack of Galois extensions; $\mathbb{Q}(\alpha, \beta) \supset \mathbb{Q}(\alpha), \mathbb{Q}(\beta) \supset \mathbb{Q}$. Therefore the Galois group consists of four elements which is generated by $\sigma$, $\eta$ which fixes $\mathbb{Q}(\alpha)$, $\mathbb{Q}(\beta)$, respectively. Note that applying elements of the Galois group on $\gamma=\alpha + \beta$, we have four different element, $\gamma$, $\sigma(\gamma)$, $\eta(\gamma)$, $\sigma(\eta(\gamma))$. Therefore, the minimal polynomial of $\alpha + \beta$ has degree four which contradicts to the fact that $\alpha+\beta$ is of degree $2$. This completes the proof.

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    $\begingroup$ How do you know that the Gal group consists of 4 elements? $\endgroup$ – Jhon Doe Oct 15 '18 at 14:25
  • $\begingroup$ @Jhon Doe $\mathbb{Q}(\alpha, \beta)$ has intermediate fields. Two successive extensions are of degree at least two. $\endgroup$ – seoneo Oct 15 '18 at 14:29
  • $\begingroup$ Oh, you mean the Galois group contains the four elements $1$, $\sigma$, $\eta$, and $\sigma\eta$, but these elements are not necessarily all the elements of the group? $\endgroup$ – user593746 Oct 15 '18 at 14:31
  • $\begingroup$ @seoneo To show that Gal group has 4 element don't you have to show that the minimal polynomials of $\alpha$ and $\beta$ are separable? $\endgroup$ – Jhon Doe Oct 15 '18 at 14:31
  • $\begingroup$ @JhonDoe I think in characteristic $0$, the minimal polynomials are always separable. $\endgroup$ – user593746 Oct 15 '18 at 14:32
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By the hypotheses, you have six rational numbers $a_0,a_1,b_0,b_1,c_0,c_1$ such that

$$ \alpha^2+a_1 \alpha +a_0=0 \tag{1} $$

$$ \beta^2+b_1 \beta +b_0=0 \tag{2} $$

$$ (\alpha+\beta)^2+c_1(\alpha+\beta) +c_0=0 \tag{3} $$

Now, (3)-(1)-(2) yields :

$$ 2\alpha\beta+(c_1-a_1)\alpha+(c_1-b_1)\beta +c_0-a_0-b_0=0 \tag{4} $$

or equivalently,

$$ \beta=\frac{-(c_1-a_1)\alpha}{2\alpha+c_1-b_1}, \alpha=\frac{-(c_1-b_1)\beta}{2\beta+c_1-a_1} \tag{5} $$

Note that the denominators are nonzero because $\alpha$ and $\beta$ have degree $2$. Clearly, (5) implies $ {\mathbb Q}(\alpha) = {\mathbb Q}(\beta)$.

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  • $\begingroup$ This is a good solution! $\endgroup$ – seoneo Nov 1 '18 at 4:07
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Here is an alternative but longer approach. Let $a(x)$, $b(x)$, and $c(x)$ be monic quadratic polynomials in $\mathbb{Q}[x]$ whose roots are $\alpha$, $\beta$, and $\gamma:=\alpha+\beta$, respectively. By the condition that $\alpha$ and $\beta$ are algebraic over $\mathbb{Q}$ with degree $2$, we see that $a$ and $b$ are irreducible over $\mathbb{Q}$. If $c$ is reducible over $\mathbb{Q}$, then $\alpha+\beta=\gamma$ is rational, so it is obvious that $\mathbb{Q}(\alpha)=\mathbb{Q}(\gamma-\alpha)=\mathbb{Q}(\beta)$. We now assume that $c$ is irreducible over $\mathbb{Q}$.

Write $p(x)$ for $a(x-\beta)\ a(x-\bar{\beta})\in\mathbb{Q}[x]$, where $\bar{\beta}$ is the other root of $b(x)$. Clearly, $p(\gamma)=0$. So, $c$ divides $p$ because $c$ is irreducible. Now, if $\bar{\alpha}$ is the other root of $a(x)$, then $$p(x)=(x-\alpha-\beta)(x-\bar{\alpha}-\beta)(x-\alpha-\bar{\beta})(x-\bar{\alpha}-\bar{\beta}).$$ Hence, there are three possibilities.

Case 1: $c(x)=(x-\alpha-\beta)(x-\bar{\alpha}-\beta)=a(x-\beta)$. Since $a$ has rational coefficients, this implies that $\beta\in\mathbb{Q}$, a contradiction.

Case 2: $c(x)=(x-\alpha-\beta)(x-\alpha-\bar{\beta})=b(x-\alpha)$. Since $b$ has rational coefficients, this implies that $\alpha\in\mathbb{Q}$, a contradiction.

Case 3: $c(x)=(x-\alpha-\beta)(x-\bar{\alpha}-\bar{\beta})$. Therefore, we have $$\alpha\bar{\alpha}+\beta{\bar{\beta}}+(\alpha\bar{\beta}+{\bar{\alpha}}\beta)=(\alpha+\beta)(\bar{\alpha}+\bar{\beta})\in\mathbb{Q}.$$ Since $\alpha\bar\alpha$ and $\beta\bar\beta$ are rational, we get $$\alpha\bar{\beta}+{\bar{\alpha}}\beta\in\mathbb{Q}.$$ Thus, $$\alpha(B-\beta)+(A-\alpha)\beta\in\mathbb{Q}$$ if $A=\alpha+\bar{\alpha}\in\mathbb{Q}$ and $B=\beta+\bar\beta\in\mathbb{Q}$. So, $\alpha\beta$ is in the rational span of $1$, $\alpha$, and $\beta$. This means $$\big[\mathbb{Q}(\alpha,\beta):\mathbb{Q}\big]\leq 3.$$ But $$\big[\mathbb{Q}(\alpha,\beta):\mathbb{Q}\big]=\big[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)\big]\ \big[\mathbb{Q}(\alpha):\mathbb{Q}\big]=2\ \big[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)\big].$$ The only even positive integer less than or equal to $3$ is $2$, so $$2\ \big[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)\big]=2$$ or $$\big[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)\big]=1.$$ This shows that $\beta\in\mathbb{Q}(\alpha)$. Thus, $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.

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