# Thread: Degree of a finite extension of towered fields.

1. ## Degree of a finite extension of towered fields.

I have a question that goes like this: Suppose that F,K,L are all fields and L contains K which contains F. If L is a finite extension of F show that L is a finite extension of K and K is a finite extension of F.

One idea I had was letting A[i]B[j] be a bases of L over F and showing we can have A[i] as a bases of L over K and B[j] as a bases of K over F but I got stuck in the algebra and think I am making heavy work of it.
Is it enough to just say that as L is a finite extension over F both F and K must be field extensions and so as L contains them they must be finite dimensional? So K must be finite dimensional over F and L must be finite dimensional over F?
Thanks

2. Prove the following:

(i) If $\displaystyle \{a_1,\ldots,a_r\}\subset L$ is linearly independent over $\displaystyle K$ , which implies $\displaystyle r\leq (L:K)$

and

$\displaystyle \{b_1,\ldots,b_s\}\subset K$ is linearly independent over $\displaystyle F$ , which implies $\displaystyle s\leq (K:F)$

then,

the $\displaystyle rs$ elements $\displaystyle a_ib_j$ of $\displaystyle L$ are linearly independent over $\displaystyle F$ .

(ii) If $\displaystyle r= (L:K), s= (K:F)$ those $\displaystyle rs$ elements of $\displaystyle L$ generate $\displaystyle L$ as a vector space over $\displaystyle F$ .

$\displaystyle (L:F)=(L:K)(K:F)$

Fernando Revilla

Edited: Sorry, I read too quickly and my hints were for a different problem, although related with it.

3. Yes this is sort of the reverse of the problem I am dealing with, I am familiar with the proof of the above but am still stuck with the original problem

4. Originally Posted by FernandoRevilla
Prove the following:

(i) If $\displaystyle \{a_1,\ldots,a_r\}\subset L$ is linearly independent over $\displaystyle K$ , which implies $\displaystyle r\leq (L:K)$

and

$\displaystyle \{b_1,\ldots,b_s\}\subset K$ is linearly independent over $\displaystyle F$ , which implies $\displaystyle s\leq (K:F)$

then,

the $\displaystyle rs$ elements $\displaystyle a_ib_j$ of $\displaystyle L$ are linearly independent over $\displaystyle F$ .

(ii) Those $\displaystyle rs$ elements of $\displaystyle L$ generate $\displaystyle L$ as a vector space over $\displaystyle F$ .

$\displaystyle (L:F)=(L:K)(K:F)$

Fernando Revilla

Edited: Sorry, I read too quickly and my hints were for a different problem, although related with it.
Do you really need that much?

You can just say $\displaystyle [L:K]$ is finite because the basis for $\displaystyle L$ as an $\displaystyle F$-vector space will also span $\displaystyle L$ as a $\displaystyle K$-vector space (it won't necessarily be a basis, but that doesn't matter).

Also, $\displaystyle [K:F]$ is finite because $\displaystyle K$ is an $\displaystyle F$-subspace of the finite dimensional $\displaystyle F$-vector space $\displaystyle L$.

5. Originally Posted by topspin1617
Do you really need that much?
Of course we don't. The proof I provided (see Lessons in Modern Algebra , Dubreil-Jacotin) allows to construct a basis of $\displaystyle L$ over $\displaystyle F$ in terms of basis of $\displaystyle L$ over $\displaystyle K$ and $\displaystyle K$ over $\displaystyle F$ .

Fernando Revilla

6. Originally Posted by kevinlightman
Yes this is sort of the reverse of the problem I am dealing with, I am familiar with the proof of the above but am still stuck with the original problem
Hint :

Following the outline (i) of my first post, analyze what would happen if $\displaystyle (L:K)=+\infty$ or $\displaystyle (K:F)=+\infty$

Fernando Revilla