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 $\displaystyle \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

$\displaystyle m(x)=m_1(x)m_2(x)m_3(x).....m_n(x)$ where $\displaystyle m_i(x) \in L[x] $ then one of $\displaystyle m_i(x) $ is the minimal polynomial of $\displaystyle \alpha $ over L and has no multiple zero since if it did so would m(x)