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Thread: Isomorphisms

  1. #1
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    Isomorphisms

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

    So here is my thoughts on this:

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

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

    The isomorphism must be of the form

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

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

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

    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 $\displaystyle \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 07:37 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 $\displaystyle \phi: \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{3})$ and look at where $\displaystyle \sqrt{2}$ maps to. Square it, and then in a few lines you can see that the thing you're mapping to must be outwith $\displaystyle \mathbb{Q}(\sqrt{3})$.

    What is it you find unsatisfactory about your proof?
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  3. #3
    Lord of certain Rings
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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 $\displaystyle \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})$

    So here is my thoughts on this:

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

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

    The isomorphism must be of the form

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

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

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

    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 $\displaystyle \mathbb{Q}(\sqrt{3}) $ but that came up empty.

    Any thoughts or comments would be awsome.

    Thanks

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

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