abstract algebra - Given $K(alpha)/K$ and $K(beta)/K$ disjoint extensions with at least one of them odd degree then $K(alpha,beta)=K(alphabeta)$

I have problems with this exercise

Let be $K(\alpha)/K$ and $K(\beta)/K$ disjoint extensions with at least one of them odd degree. Prove that $\alpha\beta$ is a primitive element for the extension $K(\alpha,\beta)/K$.

Some of my ideas were

• Prove that $K(\alpha,\beta) \subset K(\alpha\beta)$ or that $K(\alpha) \subset K(\alpha\beta)$.




• Use that in this situation $K(\alpha)=K(\alpha^2)$.




• Tried to relate the irreducible polynomials from the extensions involved.

I didn't find anything useful. Can you help me?

Thank you in advance.

Answer

This is false - a counterexample is given by $ \alpha = \sqrt[3]{2} $, $ \beta = \sqrt[3]{3} $, $ K = \mathbf Q $. The fields $ \mathbf Q(\sqrt[3]{2}) $ and $ \mathbf Q(\sqrt[3]{3}) $ intersect trivially (left as an exercise), are both of degree $ 3 $ over $ \mathbf Q $, but $ \alpha \beta = \sqrt[3]{6} $ is of degree $ 3 $ over $ \mathbf Q $, so is not a primitive element of the extension $ \mathbf Q(\alpha, \beta)/\mathbf Q $, which is of degree $ 9 $.