# Finite Extensions

• Feb 15th 2006, 02:46 PM
ThePerfectHacker
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, $F\leq E$ are fields, and $\alpha,\beta\in E$ algebraic over $F$ and $[F(\alpha):F]=[F(\beta):F]$.
Prove (or disprove):
$F(\alpha)=F(\beta)$.
???
• Feb 15th 2006, 04:08 PM
topsquark
Quote:

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

I might have the wrong idea here, because I'm not sure how you are using $\leq$, but my thought is:

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

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

-Dan
• Feb 16th 2006, 01:31 PM
ThePerfectHacker
You said well, thank you.