Let $E/F$ be a finite field extension, and $D/E$ be a field extension. Let $\alpha \in D$ be algebraic over $F$. Prove: $$ [E(\alpha) : F(\alpha)] \leq [E:F]$$
I'm not even sure why this is intuitively true. I recently did a question proving that the degree of the minimum polynomial for $\alpha$ over $E$ is less than or equal to the minimum polynomial for $\alpha$ over $F$ and I could quickly see that it made sense because $E$ is an extension of $F$, so the min poly in $F$ either reduced in $E$ and loses degree, or has the same degree. When it comes to this, I'm trying to look at the basis of $E$ as an $F$-vector space and similarly for $E(\alpha)$ and $F(\alpha)$ but nothing is clicking.
I was playing around with the facts that
- Finite extension $\implies$ algebraic $\implies$ finitely generated by elements in $F$
- $\alpha$ is algebraic in $F$ so it has a min poly in E and F
- So $E(\alpha)$ = $ E(\gamma_1, \dots, \gamma_n)(\alpha) = E(\gamma_1, \dots ,\gamma_n, \alpha)$ where $\gamma_i \in F$ … Something tells me that the $\gamma_i$ make up a basis for $E$ over $F$ but I don't know how to prove that (or if it's even true)
But I'm not even sure if trying to look at a basis is the correct way to approach the question — I'm not aware of any other slick characterizations of the degree of a field extension. I'm also unsure of the significance of $D/E$ being a field extension. I think there must be something I'm missing with regards to its significance in this set-up.
Any help is appreciated! Thanks!
Best Answer
$[E(a):F]=[E(a):E][E:F]=[E(a):F(a)][F(a):F]$ but $[E(a):E]\le[F(a):F]$, therefore....