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 !

  • $\begingroup$ you might know that $[F(\alpha,\beta):F]=[F(\alpha,\beta):F(\alpha)][F(\alpha):F]$ $\endgroup$ – user8268 Nov 18 '14 at 13:01
  • $\begingroup$ of course but it doesn't help ! $\endgroup$ – idm Nov 18 '14 at 13:26
  • $\begingroup$ OK: product of finite numbers is finite, and $[F(\alpha):F]$ is finite if $\alpha$ is algebraic over $F$ $\endgroup$ – user8268 Nov 18 '14 at 13:28

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):


and the above is finite since we have the product of two finite extensions.

Everything now follows from

$$\alpha\,,\,\alpha^{-1}\,,\,\,\alpha\pm\beta\,,\,\,\alpha\beta\in F(\alpha\,,\,\,\beta)\;$$

  • $\begingroup$ The thing is how do you know that $[F(\alpha)(\beta):F]$ is finite ? Because $\beta$ algebraic over $F$ doesn't implies that $[F(\beta):F]$ is finish ! (but the converse is true) $\endgroup$ – idm Nov 18 '14 at 13:45
  • $\begingroup$ Of course it does imply that! In fact, a specific basis can even be given...You're confusing this with the fact that there can be algebraic extensions of infintie degree (ordimension). That's another matter. $\endgroup$ – Timbuc Nov 18 '14 at 13:50

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