1. ## Finite Extensions

I was thinking about this and was not able to prove this. I believe that if this is true then it might be useful in field theory.

If, $\displaystyle F\leq E$ are fields, and $\displaystyle \alpha,\beta\in E$ algebraic over $\displaystyle F$ and $\displaystyle [F(\alpha):F]=[F(\beta):F]$.
Prove (or disprove):
$\displaystyle F(\alpha)=F(\beta)$.
???

2. Originally Posted by ThePerfectHacker
I was thinking about this and was not able to prove this. I believe that if this is true then it might be useful in field theory.

If, $\displaystyle F\leq E$ are fields, and $\displaystyle \alpha,\beta\in E$ algebraic over $\displaystyle F$ and $\displaystyle [F(\alpha):F]=[F(\beta):F]$.
Prove (or disprove):
$\displaystyle F(\alpha)=F(\beta)$.
???
I might have the wrong idea here, because I'm not sure how you are using $\displaystyle \leq$, but my thought is:

Let $\displaystyle F= \mathbb{Q}$ and $\displaystyle E= \mathbb{Q} [ \sqrt2 ,\sqrt3 ]$. $\displaystyle \sqrt2$ and $\displaystyle \sqrt3$ are both algebraic over $\displaystyle \mathbb{Q}$ and $\displaystyle [ \mathbb{Q} ( \sqrt2 ): \mathbb{Q} ] = [ \mathbb{Q} (\sqrt3 ): \mathbb{Q} ]$.

However, $\displaystyle \mathbb{Q} ( \sqrt2 ) \neq \mathbb{Q} ( \sqrt3 )$.

-Dan

3. You said well, thank you.