Adjoining elements in subextension Suppose $L/\mathbb{Q}$ is an extension of degree 10 and $\mathbb{Q} \subseteq E_1 \subseteq L$ with $[E_1: \mathbb{Q}]$ = 2 and $\mathbb{Q} \subseteq E_2 \subseteq L$ with $[E_2: \mathbb{Q}]$ = 5. Also, let $x_1 \in E_1$ and $x_2 \in E_2$. Must there exist $\alpha_1, \alpha_2 \in \mathbb{Q}$ such that $Q(\alpha_1x_1+\alpha_2x_2) = L$?
Clearly, if $x_1,x_2 \in \mathbb{Q}$ then the answer to the question is no, but what if this is not the case?
 A: Let $x_1\in E_1$, $x_2\in E_2$ and $\alpha_1,\alpha_2\in\Bbb{Q}$. Set $\alpha:=\alpha_1x_1+\alpha_2x_2$ and $K:=\Bbb{Q}(\alpha)$ and $d:=[K:\Bbb{Q}]$.
If $x_1\in\Bbb{Q}$ then $\alpha_1x_1+\alpha_2x_2\in E_2$ and hence $\Bbb{Q}(\alpha_1x_1+\alpha_2x_2)\subset E_2$, so in particular $\Bbb{Q}(\alpha_1x_1+\alpha_2x_2)\neq L$. Similarly if $x_2\in\Bbb{Q}$ then $\Bbb{Q}(\alpha_1x_1+\alpha_2x_2)\neq L$. So if $\Bbb{Q}(\alpha_1x_1+\alpha_2x_2)=L$ then necessarily $x_1,x_2\notin\Bbb{Q}$.
Now suppose $x_1,x_2\notin\Bbb{Q}$, set $\alpha:=\alpha_1x_1+\alpha_2x_2$ and $d:=[\Bbb{Q}(\alpha):\Bbb{Q}]$. Then $d\mid10$, so $d\in\{1,2,5,10\}$.
If $d=1$ then $\alpha\in\Bbb{Q}$, which implies $\alpha_1=\alpha_2=0$.
If $d=2$ then there exist $u,v\in\Bbb{Q}$ such that $\alpha^2+u\alpha+v=0$, which is equivalent to
$$\alpha_2^2x_2^2+(2\alpha_1\alpha_2x_1+u\alpha_2)x_2+\alpha_1^2x_1^2+u\alpha_1x_1+v=0.$$
If $\alpha_2\neq0$ then this shows that $x_2$ is a root of a quadratic polynomial with coefficients in $E_1$. In particular this implies $[E_1(x_2):E_1]\leq2$ and hence $[E_1(x_2):\Bbb{Q}]\leq4$. But $E_2\subset E_1(x_2)$ and $[E_2:\Bbb{Q}]=5$, a contradiction. So $d=2$ implies $\alpha_2=0$.
If $d=5$ then an entirely analogous argument shows that $\alpha_1=0$.
Hence by exhaustion, if $x_1\notin E_1$ and $x_2\notin E_2$, then for any nonzero $\alpha_1,\alpha_2\in\Bbb{Q}$ we have
$$\Bbb{Q}(\alpha_1x_1+\alpha_2x_2)=L.$$
