1) Let $E/F$ an extension and let $\alpha,\beta\in E$ be algebraic elements over $F$. If $\alpha\neq 0$, prove that $\alpha+\beta$, $\alpha\beta$ and $\alpha^{-1}$ are all algebraic over $F$.
2) If $E/F$ is an extension, prove that the subset $$K=\{\alpha\in E\mid \alpha\text{ is algebraic over }F\}$$ is a subfield of $E$ containing $F$.
For the question 1) I have for hint to show that $F(\alpha,\beta)$ is a finite vector space over $F$. I don't know how to show it. The only thing I know is that $$[F(\alpha,\beta):F(\alpha)][F(\alpha):F]=[F(\alpha,\beta):F(\beta)][F(\beta):F]$$
Then, if it's if finish, I know that $F(\alpha,\beta)\subset E$ is an algebraic extension. Then, all element of $F(\alpha,\beta)$ is algebraic. Would it be enough too conclude that $\alpha+\beta$, $\alpha\beta$ and $\alpha^{-1}$ are algebraic ? To me yes, because $F(\alpha,\beta)$ is a field and so, $\alpha+\beta$, $\alpha\beta$ and $\alpha^{-1}$ are in $F(\alpha,\beta)$ and so, they are algebraic. What do you think ?
For 2) To me it's a simple consequence of the question 1) but I'm probably wrong...
Thanks !
How doesn't that help?
$\;\beta\;$ algebraic over $\;F\;\implies\;\beta$ algebraic over $\;F(\alpha)\;$ , and thus, since $\;F(\alpha,\beta)=F(\alpha)(\beta)\;$ (you can take this as an isomorphism instead of equality if you wish):
$$[F(\alpha)(\beta):F]=[F(\alpha)(\beta):F(\beta)]\cdot[F(\beta):F]$$
and the above is finite since we have the product of two finite extensions.
Everything now follows from
$$\alpha\,,\,\alpha^{-1}\,,\,\,\alpha\pm\beta\,,\,\,\alpha\beta\in F(\alpha\,,\,\,\beta)\;$$