Determine the Degree extension of $K$ over $\mathbb{Q}$ Let $F$ be a number field and $a,b\in F[x]$ be two distinct irreducible polynomials. Suppose $\deg(a)=\deg(b)=p$ for some prime $p\in \mathbb{Z}^+$. Also suppose these polynomials have the property that if they have a root in a field $L$ then they also split $L[x]$. Let $\alpha,\beta\in \mathbb{C}$ are respective roots of $a$ and $b$ and assume $\mathbb{Q}(\alpha)\neq \mathbb{Q}(\beta)$.
I would like to show $[K:\mathbb{Q}]=p^2$ where $K=\mathbb{Q}(\alpha,\beta)$.
I am pretty sure it is true, but I am stuck. Below is how far I have gotten.
$\textbf{My attempt:}$
I know that $K$, $\mathbb{Q}(\alpha)$, and $\mathbb{Q}(\beta)$ are Galois over $\mathbb{Q}$. Also, $[\mathbb{Q}(\alpha):\mathbb{Q}]=p$. I know that if I can show $[K:\mathbb{Q}(\alpha)]=p$, then I am done. Since $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha)(\beta)$, it would suffice to show that $\min(\beta,\mathbb{Q}(\alpha))=b$.
I know that if $b$ factors into any linear terms in $\mathbb{Q}(\alpha)[x]$, then I can finish the argument, but I wondering if it is possible $b$ could factor without linear terms. For example, if $p=13$, would it be possible $b$ could factor into a quartic and ninth degree polynomial in $\mathbb{Q}(\alpha)[x]$?
 A: Note that $\Bbb Q(\alpha)$ is the splitting field of $a$ and $\Bbb Q(\beta)$ for $b$. The polynomial $ab$ splits in $\Bbb Q(\alpha,\beta)$ since $a$ and $b$ each have a root, so $\Bbb Q(\alpha, \beta)$ is the splitting field of $ab$, and thus Galois.
Let $G=\operatorname{Gal}(K/\Bbb Q)$, and let $H$ be the subgroup whose fixed field is $\Bbb Q(\alpha)$, and similarly $H'$ for $\Bbb Q(\beta)$. That is, $[G:H]=[G:H']=p$. Both $H$ and $H'$ are normal in $G$ since their corresponding fixed fields are normal extensions, so we may consider the subgroup $HH'$. We know $H\neq H'$ since $\Bbb Q(\alpha)\neq \Bbb Q(\beta)$, and since $H\cap H'$ is either $1$ or $p$, it must be $1$.
Next, we claim $HH'=G$. If $x\in K$ is in the fixed field of $HH'$, then in particular it is in the fixed field of both $H$ and $H'$, i.e., $x\in \Bbb Q(\alpha)\cap \Bbb Q(\beta)$. But each of these has prime order, so there are no nontrivial subfields. They are also not equal, so $x\in \Bbb Q$, the fixed field of $HH'$ is $\Bbb Q$, and $HH'=G$.
Finally, $|G|=|HH'|=\frac{|H||H'|}{|H\cap H'|}=p^2$, so $[K:\Bbb Q]=p^2$.
