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Math Help - Extension fields

  1. #1
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    Extension fields

    is √7 ∈ Q((5+√7)^(1/2)

    Using:

    [E:G]=[E:F][F:G]

    This gives me that the dimension of Q((5+√7)^(1/2) as a vector space over Q(√7 ) is 2.
    But i don't know if the irreducible polynomial of degree 4 over Q((5+√7)^(1/2), is still irreducible over Q(√7 ).Is it sufficient to state that a basis is [1,5+√7)^(1/2)] and
    √7 cannot be written as a linear combination of this basis?
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  2. #2
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    Quote Originally Posted by ulysses123 View Post
    is √7 ∈ Q((5+√7)^(1/2)

    Using:

    [E:G]=[E:F][F:G]

    This gives me that the dimension of Q((5+√7)^(1/2) as a vector space over Q(√7 ) is 2.


    Stop here! This already tells you that \sqrt{7}\notin\mathbb{Q}(\sqrt{5+\sqrt{7}}) , otherwise you'd

    get that dim_{\mathbb{Q}(\sqrt{7})}\mathbb{Q}(\sqrt{5+\sqrt  {7}})=1 ...

    Tonio


    But i don't know if the irreducible polynomial of degree 4 over Q((5+√7)^(1/2), is still irreducible over Q(√7 ).Is it sufficient to state that a basis is [1,5+√7)^(1/2)] and
    √7 cannot be written as a linear combination of this basis?
    .
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  3. #3
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    That doesn't seem like its sufficient,

    as a counter example take √3 and Q(√(2))

    Using [E:G]=[E:F][F:G]
    where E=Q(√(2))
    F=Q(√(3))
    G=Q
    the dimension of [Q(√(2)):Q(√(3))]=1
    But still √(3) is not actually contained in Q(√(2)

    Since [Q(√(2))(√(3)):Q]=4
    and each of root 2 and root 3 have dimension 2 over Q, this then says that there is no unique way to write root 3 in Q(√(2), otherwise if there was the dimension of [Q(√(2))(√(3)):Q]would be two.
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  4. #4
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    Quote Originally Posted by ulysses123 View Post
    That doesn't seem like its sufficient,

    as a counter example take √3 and Q(√(2))

    Using [E:G]=[E:F][F:G]
    where E=Q(√(2))
    F=Q(√(3))
    G=Q
    the dimension of [Q(√(2)):Q(√(3))]=1


    Uh? How exactly is \mathbb{Q}(\sqrt{2}} a vector space over \mathbb{Q}(\sqrt{3})??
    Either one is a subfield of the

    other, and the multiplication of vectors by scalars is obvious, or else you must define this product.

    Tonio



    But still √(3) is not actually contained in Q(√(2)

    Since [Q(√(2))(√(3)):Q]=4
    and each of root 2 and root 3 have dimension 2 over Q, this then says that there is no unique way to write root 3 in Q(√(2), otherwise if there was the dimension of [Q(√(2))(√(3)):Q]would be two.
    .
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