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?


3 Answers 3


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)$.

  • $\begingroup$ This is a good solution! $\endgroup$
    – seoneo
    Nov 1, 2018 at 4:07

We feel like there should be some correction to the following solution. However, because a good time have past since then (a few years), we decided to write a completely rewritten answer.

Suppose that the numbers $\alpha$, $\beta$, $\alpha+\beta$ are algebraic over $\mathbb{Q}$ with algebraic degree $2$ and $\mathbb{Q}(\alpha) \neq \mathbb{Q}(\beta)$. Since $K=\mathbb{Q}(\alpha,\beta)$ is the splliting field of the product of the irreducible polynomials of $\alpha$ and $\beta$, it is normal and hence, Galois. Regarding the tower $K-\mathbb{Q}(\alpha)-\mathbb{Q}$, $[K:\mathbb{Q}]=4$ and hence the Galois group is of order $4$ and abelian.

The Galois group $G=\mathrm{G}(K/\mathbb{Q})$ has two elements, say $\sigma$ and $\tau$ which fixes $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$, respectively and generate $G$. Note that $\sigma$ fixed the conjugate $\overline{\alpha}$ also and since $\tau(\alpha) = \overline{\alpha}$. The similar fact holds for $\beta$.

Since $\tau \sigma \neq 1, \tau, \sigma$, we have $G = \left\{ 1,\tau, \sigma , \tau \sigma =\sigma \tau \right\}$. Let $\eta \in G$ be such that $<\eta> = \mathrm{G} (K/\mathbb{Q}[\gamma])$. Then $\eta$ is of order two and $\eta \neq \tau$, $\eta \neq \sigma$. Hence $\eta = \sigma \tau$. But then $\eta (\gamma) = \sigma \tau (\alpha + \beta) = \overline{\alpha} + \overline{\beta}$, so $\eta(\gamma) = \gamma$ implies $\sigma(\beta) + \tau(\alpha) = \alpha + \beta$.

Now $\tau(\gamma)$ and $\sigma(\gamma)$ are both the conjugate of $\gamma$ and $\gamma$ is of degree $2$. Hence they are same. Therefore, from $\tau(\alpha + \beta) = \sigma(\alpha+\beta)$, we have $\tau(\alpha)-\sigma(\beta) = \alpha - \beta$, and together with $\sigma(\beta) + \tau(\alpha) =\alpha+\beta$, $\tau(\alpha) = \alpha$ which is an absurdity. Hence $\mathbb{Q}(\alpha) = \mathbb{Q} (\beta)$이다.

Below is the former answer given a few years ago.

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.

  • 2
    $\begingroup$ How do you know that the Gal group consists of 4 elements? $\endgroup$
    – Jhon Doe
    Oct 15, 2018 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, 2018 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, 2018 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, 2018 at 14:31
  • $\begingroup$ @JhonDoe I think in characteristic $0$, the minimal polynomials are always separable. $\endgroup$
    – user593746
    Oct 15, 2018 at 14:32

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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