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Math Help - Galois groups

  1. #1
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    Galois groups

    If L:K (field extension) is finite and seperable but not normal, then
    |Gal(L:K)|<[L:K]. where Gal(L:K) is the group of k-Auts on L. [L:K] is the degree of the extension.

    Since L:K is finite and seperable it is simple so i can say L=K(\alpha).
    I think i need to show that m(x) the minimal polynomial of \alpha cannot split over L.
    All i can say is that since it's not seperable, there is an irreducible polynomial in K[x] which doesn't split over L. how can i show that m(x) cannot split also.
    Any help is good.
    Thanks

    EDIT: also i know [L:K]=degree of m(x) and |Gal(L:K)|=the number of zeroes m(x) has in L

    EDIT2:Hang on can't i use the definition of normality and say; it's not normal so there is an irreducible poly f(x) over K[x] which has a zero in L=K(\alpha) but doesn't split over L. so f(\alpha)=0 and so m(x)|f(x). and m(x)=f(x). so m(x) doesn't split???? i'm a bit wary of the f(\alpha)=0 part
    Last edited by Krahl; March 14th 2010 at 06:37 PM.
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  2. #2
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    Quote Originally Posted by Krahl View Post
    If L:K (field extension) is finite and seperable but not normal, then
    |Gal(L:K)|<[L:K]. where Gal(L:K) is the group of k-Auts on L. [L:K] is the degree of the extension.

    Since L:K is finite and seperable it is simple so i can say L=K(\alpha).
    I think i need to show that m(x) the minimal polynomial of \alpha cannot split over L.
    All i can say is that since it's not seperable, there is an irreducible polynomial in K[x] which doesn't split over L. how can i show that m(x) cannot split also.
    Any help is good.
    Thanks

    EDIT: also i know [L:K]=degree of m(x) and |Gal(L:K)|=the number of zeroes m(x) has in L
    Let K_1 be a fixed field of Gal(L:K). It follows that K \subset K_1 \subset L. Then [L:K]=|Gal(L:K)||K_1:K|. Thus, |Gal(L:K)| < [L:K] (It never equals because L is not the normal extension over K).

    For example, let K=\mathbb{Q} and L=\mathbb{Q}(\sqrt[3]{2}). Then L is the finite and separable extension over K but not the normal extension over K. You see that [L:K]=3, |Gal(L:K)|=1, |K_1:K|=3.
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  3. #3
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    hi thanks for your reply. what does |K':K| mean with the vertical bars.
    but surely if <br />
[L:K]=3, |Gal(L:K)|=1, |K_1:K|=3<br />
is true then L=K_1 since [L:K]=[L:K_1][K_1:K]
    by the tower law.
    also how can K_1 be a fixed field of a group.
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  4. #4
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    Quote Originally Posted by Krahl View Post
    hi thanks for your reply. what does |K':K| mean with the vertical bars.
    but surely if <br />
[L:K]=3, |Gal(L:K)|=1, |K_1:K|=3<br />
is true then L=K_1 since [L:K]=[L:K_1][K_1:K]
    by the tower law.
    also how can K_1 be a fixed field of a group.
    A typo there. [K_1:K] instead of |K_1:K|.
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  5. #5
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    Well i managed to do it i think for anyone interested;

    Proof

    L:K \; finite \; and \; separable \Longrightarrow it's \; simple \Longrightarrow L=K(\alpha)

    L:K \; finite \Longrightarrow algebraic

    so let m(x) be the minimum polynomial of \alpha over K.

    if deg(m(x))=1 then \alpha \in K contradiction since K:K is normal.

    so deg(m(x))>1, say n.

    suppose m(x) splits in L.
    m(x)=(x-\alpha_1)...(x-\alpha_n) and \alpha_1=\alpha

    let M be a splittin field for m(x) over K such that M is a subset of L. then we must have that \alpha_i \in M for all i. and M is a field so K(\alpha) \subset M

    so L is a splitting field for m(x) over K. but then this means L:K is finite and normal. so m(x) cannot split in L (or any subset of L). So

    |Gal(L:K)|= no. \; of \; zeros \; of \; m(x) \; in \; L < deg(m(x))=[L:K]
    Last edited by Krahl; April 19th 2010 at 05:58 PM.
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