1. ## finite, separable

Hi
I'm trying to prove the following theorem.

Theorem;
Suppose K,L,M are fields with K subset of L subset of M such that M:K (field extension) is algebraic and separable. Then both L:K and M:L are algebraic and seperable.

Proof:
If M:K is algebraic then it is finite. since M:K is finite then tower law says that both L:K and M:L are finite and hence algebraic. Since M:K is seperable every element of M is seperable over K. So the minimal polynomial of every element of M is seperable over K. Since L is a subset of M, every element of L is seperable over K also. so L:K is seperable.

If i can show that the minimal polynomials over K of these elements of M are also irreducible over L then i'm done, since then these minimal polynomials will still have no multiple zero.

I need help in showing M:L is seperable. Also is what i have done so far correct?

Any help appreciated thanks.

EDIT: Actually i don't think "If i can show that the minimal polynomials over K of these elements of M are also irreducible over L" is always true.

EDIT2: I wonder if ican do the following;
Suppose $\alpha \in M$ and m(x) is it's minimal polynomial over K[x]. If m(x) is irreducible over L[x] then all's good. if not then
$m(x)=m_1(x)m_2(x)m_3(x).....m_n(x)$ where $m_i(x) \in L[x]$ then one of $m_i(x)$ is the minimal polynomial of $\alpha$ over L and has no multiple zero since if it did so would m(x)

2. Hi

If M:K is algebraic then it is finite
The converse is true, but for instance the algebraic closure of a prime field ( $\mathbb{F}_p\ \text{or}\ \mathbb{Q}$) is an algebraic non finite extension.

Your edit 2 is good, perhaps just use that given an element in $M,$ its minimal polynomial over $L$ divides its minimal polynomial over $K$ (does not need the factoriality of $L[X]$)

3. Agreed, i think i was thinking of splitting fields.

So maybe i should say If every element of M is algebraic over K then since L is a subset of M, every element of L is algebraic over K. Also every element of M is algebraic over K and so they must also be algebraic over L since K[x] is a subset of L[x].

so it's actually simpler.