# Degree of field extensions in $\mathbb{Q}$ with two algebraic elements

I've got a question regarding the degree of field extensions.

The question is as follows: given two algebraic elements $\alpha, \beta \in \mathbb{Q}$, assume that $[\mathbb{Q}(\alpha):\mathbb{Q}]$ and $[\mathbb{Q}(\beta):\mathbb{Q}]$ are relatively prime. Show that $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}] = [\mathbb{Q}(\alpha):\mathbb{Q}]\cdot [\mathbb{Q}(\beta):\mathbb{Q}]$.

We know that both $[\mathbb{Q}(\alpha):\mathbb{Q}]$ and $[\mathbb{Q}(\beta):\mathbb{Q}]$ are finite because $\alpha$ and $\beta$ are algebraic, so we could say $[\mathbb{Q}(\alpha):\mathbb{Q}] = n$ and $[\mathbb{Q}(\beta):\mathbb{Q}] =m$. We also know that $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}]$ is finite. I'm assuming I have to use that given a field $K$ and two field extensions $L,M$ where $K \subset L \subset M$, we have $[M:K]=[M:L]\cdot[L:K]$. But I'm not sure what to do next, because I don't know how to show that these extensions match these criteria.

• You have not used the assumption about $n$ and $m$ being coprime. That is absolutely crucial here! And undoubtedly has been handled on our site many times over. Recall that $n=[\Bbb{Q}(\alpha):\Bbb{Q}]$ is a factor of $[\Bbb{Q}(\alpha,\beta):\Bbb{Q}]$. Same for $m$. – Jyrki Lahtonen Dec 17 '17 at 14:46

• $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}]$ is a common multiple of $[\mathbb{Q}(\alpha):\mathbb{Q}]$ and $[\mathbb{Q}(\beta):\mathbb{Q}]$
• $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}(\alpha)] \le [\mathbb{Q}(\beta):\mathbb{Q}]$