Find the degree over a field

Question

Let $K$ be a field and let $\alpha, \beta \in \overline{K}$. Suppose that $m,n \in \mathbb{N}$ are the lowest natural numbers such that $\alpha^n \in K$ and $\beta^m \in K$. It is now easy to see that the extension $$K \subset K\left( \alpha \right) $$ has degree $n$, because $\alpha^n \in \mathbb{Q}$. But I don't know how to find the degree of the extension $$K\left(\alpha \right) \subset K\left(\alpha , \beta \right). $$ Is there a general way of solving this?

Answer 1

It is not easy to show that the degree of $\alpha$ over $K$ is $n$, because it can be less than $n$.

For instance, consider $\omega=\cos(2\pi/3)+i\sin(2\pi/3)$, which is algebraic over $\mathbb{Q}$. Then $\omega,\omega^2\notin\mathbb{Q}$ and $\omega^3\in\mathbb{Q}$, but the degree of $\omega$ over $\mathbb{Q}$ is $2$, because its minimal polynomial is $X^2+X+1$.

What you can say is that $[K(\alpha,\beta):K]\le mn$, but not much more.

Continuing with the example above, if $\xi$ is a primitive sixth root of $1$, then the minimal integer such that $\xi^n\in\mathbb{Q}$ is $6$; however the degree of $\mathbb{Q}(\omega,\xi)$ is $2$.

Answer 2

I think the best you can say is that the degree is at most $\;m\;$ , but nothing beyond this.

For example, if

$$K=\Bbb Q\;,\;\;\alpha=\sqrt2\,,\;\;\sqrt[4]3\;,\;\;\text{then}\;\;[\Bbb Q(\sqrt2)(\sqrt[4]3):\Bbb Q(\sqrt2)]=4$$

since $\;x^4-3\;$ irreducible in $\;\Bbb Q(\sqrt2)[x]\;$, but on the other side

$$K=\Bbb Q\;,\;\;\alpha=\sqrt2\,,\;\;\sqrt[4]2\;,\;\;\text{then}\;[\Bbb Q(\sqrt2)(\sqrt[4]2):\Bbb Q(\sqrt2)]=2$$

since $\;\sqrt[4]2\;$ is a root of $\;x^2-\sqrt2\in\Bbb Q(\sqrt2)[x]\;$