# Product of degree of two field extensions of prime degree

Let $L/K$ be a field extension. Let be $\alpha, \beta \in \mathbb{C}$, such that $[\mathbb{Q}(\alpha):\mathbb{Q}] = p$, and $[\mathbb{Q}(\beta):\mathbb{Q}] = q$, for some prime numbers $p$ and $q$. Assume $p \neq q$. Prove that:

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\mathbb{Q}(\alpha,\beta):Q] = [\mathbb{Q}(\alpha):\mathbb{Q}] \cdot [\mathbb{Q}(\beta):\mathbb{Q}]$

I have no idea how to proceed on this, besides proving that the left hand side is $pq$.

You have $\mathbb{Q}\subset \mathbb{Q}(\alpha) \subset \mathbb{Q}(\alpha,\beta)$, so $$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}],$$which means $p=[\mathbb{Q}(\alpha):\mathbb{Q}]$ divides $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$. The same argument applies if you replace $\alpha$ with $\beta$, so $q=[\mathbb{Q}(\beta):\mathbb{Q}]$ divides $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$. Since $p,q$ are distinct primes, $pq$ divides $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]$.

Moreover,$$[\mathbb{Q}(\alpha)(\beta):\mathbb{Q}(\alpha)]\leq [\mathbb{Q}(\beta):\mathbb{Q}]=q$$because the minimal polynomial of $\beta$ over $\mathbb{Q}$ is also a polynomial in $\mathbb{Q}(\alpha)$. Thus $$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]\leq pq.$$

Thus we conclude that $$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=pq=[\mathbb{Q}(\alpha):\mathbb{Q}]\cdot[\mathbb{Q}(\beta):\mathbb{Q}].$$

• The same argument works if $p$ and $q$ are merely coprime. – lhf Jan 16 '16 at 19:13
• Can you please clarify why the minimal polynomial of $\beta$ over $\mathbb{Q}$ is also a polynomial in $\mathbb{Q}(\alpha)$? And how does it follow from this that $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}(\alpha)]\leq [\mathbb{Q}(\beta):\mathbb{Q}]=q$? – sequence Mar 24 '17 at 15:25
• @sequence $\Bbb Q\subset \Bbb Q(\alpha)$, therefore any polynomial with coefficients in $\Bbb Q$ can also be seen as a polynomial with coefficients in $\Bbb Q(\alpha)$. Thus the minimal polynomial of $\beta$ over $\Bbb Q(\alpha)$ must divide the minimal polynomial of $\beta$ over $\Bbb Q$, and thus have lower degree. – Arnaud D. Mar 24 '17 at 15:42