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Math Help - Determining Subfields

  1. #1
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    Determining Subfields

    (a) Determine all subfields of \mathbb{Q}(\sqrt{7}). (b) Prove that \mathbb{Q}(\sqrt{5}) \ncong \mathbb{Q}(\sqrt{-3}).

    Attempt:
    For the first part I think \mathbb{Q} (\sqrt{7})= \{ a+b\sqrt{7} ; a,b \in \mathbb{Q} \}. So, I know about the "Subfield Test", but how would I find the possible subfields to be tested? Or is there a simpler way of dealing with this problem?

    For the second part to show they're not isomorphic do I just need to show there is not a one to one correspondance? I think \mathbb{Q} (\sqrt{-3}) has complex numbers whereas the other one does not...
    Last edited by demode; October 12th 2010 at 04:22 AM.
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  2. #2
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    Quote Originally Posted by demode View Post
    (a) Determine all subfields of \mathbb{Q}(\sqrt{7}). (b) Prove that \mathbb{Q}(\sqrt{5}) \ncong \mathbb{Q}(\sqrt{-3}).

    Attempt:
    For the first part I think \mathbb{Q} (\sqrt{7})= \{ a+b\sqrt{7} ; a,b \in \mathbb{Q} \}. So, I know about the "Subfield Test", but how would I find the possible subfields to be tested? Or is there a simpler way of dealing with this problem?

    For the second part to show they're not isomorphic do I just need to show there is not a one to one correspondance? I think \mathbb{Q} (\sqrt{-3}) has complex numbers whereas the other one does not...


    As linear spaces, \dim_\mathbb{Q}\left(\mathbb{Q}(\sqrt{7}\right)=2 and since any subfield would be a linear subspace of

    the last one, there are no many subfields left out there, right?

    About the non-isomorphism: you must know by now that two quadratic extensions of a field are isomorphic

    iff they're generated by elements hich are roots of the same irreducible quadratic pol. over the field, so...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    As linear spaces, \dim_\mathbb{Q}\left(\mathbb{Q}(\sqrt{7}\right)=2 and since any subfield would be a linear subspace of

    the last one, there are no many subfields left out there, right?
    So since the dimension is 2, there are 2 elements in its basis. But the question doesn't ask how many there are, it's asking to identify them! Do I somehow need to find two elements in \mathbb{Q} such that \mathbb{Q}(\sqrt{7}) is spanned by them?

    About the non-isomorphism: you must know by now that two quadratic extensions of a field are isomorphic

    iff they're generated by elements hich are roots of the same irreducible quadratic pol. over the field, so...

    Tonio
    So I must show that generators of \mathbb{Q}(\sqrt{-3}) and \mathbb{Q}(\sqrt{5}) are not the roots of the same polynomial. How do I find them? I'm very confused...

    Here's an alternative approach:

    Suppose that \phi: \mathbb{Q}(\sqrt{-3}) \to \mathbb{Q}(\sqrt{5}) is an isomorphism. Since \phi(1) = 1 (is this right?), we have \phi(-3)=-3. Then

    -3=\phi(-3)=\phi(\sqrt{-3} \sqrt{-3})= (\phi(\sqrt{-3}))^2

    This is impossible, since \phi(\sqrt{-3}) is a real number.

    Is this correct?
    Last edited by demode; October 12th 2010 at 02:53 PM.
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    Quote Originally Posted by demode View Post
    So since the dimension is 2, there are 2 elements in its basis. But the question doesn't ask how many there are, it's asking to identify them! Do I somehow need to find two elements in \mathbb{Q} such that \mathbb{Q}(\sqrt{7}) is spanned by them?



    So I must show that generators of \mathbb{Q}(\sqrt{-3}) and \mathbb{Q}(\sqrt{5}) are not the roots of the same polynomial. How do I find them? I'm very confused...

    Here's an alternative approach:

    Suppose that \phi: \mathbb{Q}(\sqrt{-3}) \to \mathbb{Q}(\sqrt{5}) is an isomorphism. Since \phi(1) = 1 (is this right?), we have \phi(-3)=-3. Then

    -3=\phi(-3)=\phi(\sqrt{-3} \sqrt{-3})= (\phi(\sqrt{-3}))^2

    This is impossible, since \phi(\sqrt{-3}) is a real number.

    Is this correct?


    Yes.

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    Yes.

    Tonio
    Do you mean my answer to part (b) is correct? Then what about part (a)? I'm really stuck on part this part...
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    Quote Originally Posted by demode View Post
    Do you mean my answer to part (b) is correct? Then what about part (a)? I'm really stuck on part this part...

    The ONLY part you did is (b), so what did I mean by "yes"?

    Anyway, what kind of subspaces has a vector space of dimension 2? Then...?

    You must have your linear algebra sharp as a butcher's knife BEFORE you want to mess with fields, extensions

    and stuff, let alone Galois Theory.

    Tonio
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    Quote Originally Posted by tonio View Post
    The ONLY part you did is (b), so what did I mean by "yes"?

    Anyway, what kind of subspaces has a vector space of dimension 2? Then...?

    You must have your linear algebra sharp as a butcher's knife BEFORE you want to mess with fields, extensions

    and stuff, let alone Galois Theory.

    Tonio
    for any subspace v, the dimension is defined to be the number of elements in a basis for v. Subspaces with dimension 2 have 2 elements in their basis (for example a plane through the origin has dim=2).

    So, are the two subfields \mathbb{Q} and \mathbb{Q}(\sqrt{7})?
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  8. #8
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    Quote Originally Posted by demode View Post
    for any subspace v, the dimension is defined to be the number of elements in a basis for v. Subspaces with dimension 2 have 2 elements in their basis (for example a plane through the origin has dim=2).

    So, are the two subfields \mathbb{Q} and \mathbb{Q}(\sqrt{7})?

    Yes...

    Tonio
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