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Math Help - Isomorphisms

  1. #1
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    Isomorphisms

    I need to prove that \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})

    So here is my thoughts on this:

    Since \mathbb{Q} is the prime subfield is must be fixed by any isomorphism from \tau : \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})

    Since \{1, \sqrt{2} \} is a basis for \mathbb{Q}(\sqrt{2}) and \{1, \sqrt{3} \} is a basis for \mathbb{Q}(\sqrt{3})

    The isomorphism must be of the form

    \tau(a+b\sqrt{2})=a+b{\sqrt{3}} or \tau(a+b\sqrt{2})=a-b{\sqrt{3}}

    Both of these only work for the addative group structure, but don't preserve multiplication. So they are not isomorphisms.

    Therefore \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})

    I feel unsatisfied with the above proof. Another thought that I had was since \mathbb{Q}(\sqrt{2}) has only two automorphism showing that neither of those could be made into an isomorphism to \mathbb{Q}(\sqrt{3}) but that came up empty.

    Any thoughts or comments would be awsome.

    Thanks

    TES
    Last edited by TheEmptySet; May 18th 2009 at 08:37 AM. Reason: Latex error :(
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  2. #2
    MHF Contributor Swlabr's Avatar
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    This came up in a thread a couple of days ago, I believe. Take \phi: \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{3}) and look at where \sqrt{2} maps to. Square it, and then in a few lines you can see that the thing you're mapping to must be outwith \mathbb{Q}(\sqrt{3}).

    What is it you find unsatisfactory about your proof?
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  3. #3
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    sah_mat has a solution to the problem:Try this
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  4. #4
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    Thanks I have something similar. I guess the reason I feel unsatified is the proof is out of place in my text. In Dummit and Foote this is in the beginning of chapter 14(the beginning of Galois theory), but I didn't use anything from the current chapter. I used only tools from the previous chapter 13. Of course this is not a reason to think a proof is incorrect, but it does make me suspect that I may have missed an applicaiton of a more recent idea.
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    I need to prove that \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})

    So here is my thoughts on this:

    Since \mathbb{Q} is the prime subfield is must be fixed by any isomorphism from \tau : \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})

    Since \{1, \sqrt{2} \} is a basis for \mathbb{Q}(\sqrt{2}) and \{1, \sqrt{3} \} is a basis for \mathbb{Q}(\sqrt{3})

    The isomorphism must be of the form

    \tau(a+b\sqrt{2})=a+b{\sqrt{3}} or \tau(a+b\sqrt{2})=a-b{\sqrt{3}}

    Both of these only work for the addative group structure, but don't preserve multiplication. So they are not isomorphisms.

    Therefore \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})

    I feel unsatisfied with the above proof. Another thought that I had was since \mathbb{Q}(\sqrt{2}) has only two automorphism showing that neither of those could be made into an isomorphism to \mathbb{Q}(\sqrt{3}) but that came up empty.

    Any thoughts or comments would be awsome.

    Thanks

    TES
    It is not hard to show \tau (r) = r, r\in \mathbb{Q}.

    Notice that \tau(\sqrt{2})^2 = \tau(\sqrt{2}^2) = \tau(2) = 2. Therefore, \tau(\sqrt{2}) = \pm \sqrt{2}.
    This means, \tau(a+b\sqrt{2}) = a \pm b\sqrt{2}.
    Now argue that \sqrt{3} \not = c + d\sqrt{2} for any c,d\in \mathbb{Q}.
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