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Math Help - finite, separable

  1. #1
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    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)
    Last edited by Krahl; March 1st 2010 at 02:56 PM.
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  2. #2
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    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])
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  3. #3
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    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.
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