# $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$ if $\alpha, \beta$ are algebraic over $\mathbb{Q}$ of degree $2$ and $\alpha +\beta$ is a root of a quadratic.

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?

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 $${\mathbb Q}(\alpha) = {\mathbb Q}(\beta)$$.

• This is a good solution! 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.

• How do you know that the Gal group consists of 4 elements? Oct 15, 2018 at 14:25
• @Jhon Doe $\mathbb{Q}(\alpha, \beta)$ has intermediate fields. Two successive extensions are of degree at least two. Oct 15, 2018 at 14:29
• 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?
– user593746
Oct 15, 2018 at 14:31
• @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? Oct 15, 2018 at 14:31
• @JhonDoe I think in characteristic $0$, the minimal polynomials are always separable.
– 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, $$\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$$.