For coprime integers $p,q>0$ and $u,v$ real numbers algebraic over $\mathbb{Q}$ of degrees $p$ & $q$, prove $[\mathbb{Q}(u,v) : \mathbb{Q}] = pq$

Question

I know that $\mathbb{Q}(u,v)$ extends $\mathbb{Q}(u)$ which extends $\mathbb{Q}$ so I was thinking I could use that $[\mathbb{Q}(u,v):\mathbb{Q}] = [\mathbb{Q}(u,v):\mathbb{Q}(u)][\mathbb{Q}(u):\mathbb{Q}]$ and calculate each of the components on the RHS, but I'm not quite sure how to do that. I suspect that $[\mathbb{Q}(u,v):\mathbb{Q}(u)] = q$ and $[\mathbb{Q}(u):\mathbb{Q}] = p$

Answer 1

By definition of degree of algebraic number, you have $[\Bbb Q(u):\Bbb Q]=p$, so that's done. You've proven that $p$ divides $[\Bbb Q(u,v):\Bbb Q]$.

Now, to finish your proof that $[\Bbb Q(u,v):\Bbb Q]=pq$, consider that $\Bbb Q(u,v)$ isn't just an extension of $\Bbb Q(u)$, but it's also an extension of $\Bbb Q(v)$, and by a similar argument as you already have, $q$ divides $[\Bbb Q(u,v):\Bbb Q]$. Now finish by using

• $[\Bbb Q(u,v):\Bbb Q(v)]\leq p\implies [\Bbb Q(u,v):\Bbb Q]\leq pq$
• $p,q$ are coprime