0
$\begingroup$

Let $E/F$ be an extention field. Then,

$$[E:F]<\infty \iff \exists a_1,...,a_n\in E \text{ algebraic over } F \text{ such that } E=F(a_1,...,a_n).$$

Proof. $(\Longleftarrow)$ Let $E=F(a_1,...,a_n)$, where $a_1,...,a_n\in E$ are algebraic over $F$. Then we have $$F\leq F(a_1) \leq F(a_1,a_2) \leq ... \leq F(a_1,...,a_{n-2})\leq F(a_1,...,a_{n-1}) \leq E=F(a_1,...,a_{n}) $$ so, from Tower Law:

$$[E:F]=[E:F(a_1,...,a_{n-1})]\cdot[F(a_1,...,a_{n-1}):F(a_1,...,a_{n-2})] \dots \\ \cdots [F(a_1,a_2):F(a_1)]\cdot[F(a_1):F] $$

Now, we observe that:

  • $a_i\in E$ is algebraic over $F \implies [F(a_i):F]=\deg Irr_{a_i,F} <\infty,\ \forall i=1,...,n.$
  • $a_2\in E$ is algebraic over $F\subseteq F(a_2) \implies a_2\in E$ is algebraic over $F(a_2) \implies [F(a_1)(a_2):F(a_1)]=\deg Irr_{a_2,F(a_1)} $ and we know that if $$F\leq F(a_1)\leq F(a_1,a_2)=F(a_1)(a_2) $$ then $Irr_{a_2,F(a_1)} \mid Irr_{a_2,F} \implies \deg Irr_{a_2,F(a_1)} =[F(a_1,a_2):F(a_1)] \leq \deg Irr_{a_2,F}<\infty$

  • $a_3 \in E$ is algebraic over $F\subseteq F(a_1,a_2) \implies a_3 \in E$ is algebraic over $F(a_1,a_2) \implies [F(a_1,a_2)(a_3):F(a_1,a_2)]=\deg Irr_{a_3,F(a_1,a_2)}$ and since $$F\leq F(a_1) \leq F(a_1,a_2) \leq F(a_1,a_2,a_3)=F(a_1,a_2)(a_3)$$ we have that $Irr_{a_3,F(a_1,a_2)} \mid Irr_{a_3,F} \implies \deg Irr_{a_3,F(a_1,a_2)} = [F(a_1,a_2,a_3):F(a_1,a_2)]< \deg Irr_{a_3,F} < \infty $

  • Working similarly for $a_n\in E$ we have that $[E:F(a_1,...,a_{n-1})]<\infty$.

So, from Tower Law, $[E:F]<\infty$. Is this right?

$(\Longrightarrow)$ Now, we assume that $[E:F]<\infty$ and that the set $$B:=\{\beta_1,...,\beta_n \} \subseteq E$$ is basis of the $F-$vector space $E$. My thought is to proove that $F(\beta_1,...,\beta_n)=\langle B \rangle $. But I stack there. Can we continue from this? And is there another way to proove this direction?

Thank you for your time.

$\endgroup$
  • $\begingroup$ For the second Part: What do you need to check for proving $E=F(B)$? And what do you have, if $B$ is a Basis? It shouldn't be too hard. $\endgroup$ – Verdruss Aug 10 '17 at 13:15
  • $\begingroup$ "$[F(\alpha):F] < \infty$" is the same as "$[F[\alpha,\alpha^{-1}]:F] < \infty$" which is the same as "$\{\alpha^n, n \in \mathbb{Z}\}$ is not a $F$-linearly independent set" which is the same as "$\alpha$ is algebraic over $F$". $\endgroup$ – reuns Aug 10 '17 at 13:15
  • $\begingroup$ @Verdruss I'm sorry, my question was: Why $F(B)=\langle B \rangle$. I fixed it. $\endgroup$ – Chris Aug 10 '17 at 13:19
  • 1
    $\begingroup$ Yes, that's the thing i wrote about in me first comment since $E=\langle B \rangle$. $\endgroup$ – Verdruss Aug 10 '17 at 13:20
  • 1
    $\begingroup$ Actually, i don't know why you do it so complicated. If any $a_i $ is algebraic over $F$, it is algebraic over any field extension. So an easy inductive argument should give you the other direction. Maybe that's what you wanted to do, but i think one could at least write it down more compact. $\endgroup$ – Verdruss Aug 10 '17 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.