TITLE: $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$ if $\alpha, \beta$ are algebraic over $\mathbb{Q}$ of degree $2$ and $\alpha +\beta$ is a root of a quadratic.
QUESTION [8 upvotes]: 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?

REPLY [4 votes]: 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, $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.