Finite Extensions

**ThePerfectHacker** · February 15th 2006, 01:46 PM

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):
$F(\alpha)=F(\beta)$ .
???

**topsquark** · February 15th 2006, 03:08 PM

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

**ThePerfectHacker** · February 16th 2006, 12:31 PM

You said well, thank you.

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