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

and

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

then,

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

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

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

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

and

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

then,

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

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

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

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

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

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 $L$ over $F$ in terms of basis of $L$ over $K$ and $K$ over $F$ .

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 $(L:K)=+\infty$ or $(K:F)=+\infty$

Fernando Revilla