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Math Help - ring isomorphism proof,need help for my homework

  1. #1
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    ring isomorphism proof,need help for my homework

    How can i prove that Q[_{\sqrt{2}}] is a ring isomorphism to Q[_{\sqrt{3}}]
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by canuhelp View Post
    How can i prove that Q[_{\sqrt{2}}] is a ring isomorphism to Q[_{\sqrt{3}}]
    Hi canuhelp.

    Try the mapping f\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3), f\left(a+b\sqrt2\right)\ \to\ a+b\sqrt3.
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  3. #3
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    Quote Originally Posted by canuhelp View Post
    How can i prove that Q[_{\sqrt{2}}] is a ring isomorphism to Q[_{\sqrt{3}}]
    Quote Originally Posted by TheAbstractionist View Post
    Hi canuhelp.

    Try the mapping f\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3), f\left(a+b\sqrt2\right)\ \to\ a+b\sqrt3.
    I disagree with TheAbstractionist because \mathbb{Q}(\sqrt{2}) is not isomorphic to \mathbb{Q}(\sqrt{3}) as fields (so they cannot be isomorphic as fields). Now since \sqrt{2},\sqrt{3} are algebraic integers it means \mathbb{Q}[\sqrt{2}] = \mathbb{Q}(\sqrt{2}),\mathbb{Q}[\sqrt{3}] = \mathbb{Q}(\sqrt{3}). Thus, asking for a ring isomorphism between \mathbb{Q}[\sqrt{2}] and \mathbb{Q}[\sqrt{3}] is asking for an isomorphism between \mathbb{Q}(\sqrt{2}) and \mathbb{Q}(\sqrt{3}) which does not exist.

    (They happen to be isomorphic as vector spaces, maybe that is what the poster was really asking).
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  4. #4
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    Construct a map \phi:\mathbb{Q}[\sqrt{2}]\rightarrow\mathbb{Q}[\sqrt{3}] such that: \phi(\sqrt{2})=\sqrt{3} and \phi(q)=q, for all q\in\mathbb{Q}.
    Then check that \phi is bijective and is a homomorphism, to show that \phi is an isomorphism.
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Hi ThePerfectHacker.

    Thanks for pointing this out. I wasnt thinking clearly when I made my post.

    Quote Originally Posted by ThePerfectHacker View Post
    (They happen to be isomorphic as vector spaces, maybe that is what the poster was really asking).
    That was probably what I had in mind.
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  6. #6
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    i tried but i am not sure it is true i wish somebody will check the solution.

    Let \phi:Q[\sqrt{2}]->Q[\sqrt{3}] be a supposed ring isomorphism. Since  \phi(x)=\phi(1*x)=\phi(1)*\phi(x)and \phi is onto, we must have \phi(1)=1. This implies that \phi(2)=2. Let \phi(\sqrt{2})=a+b*\sqrt{3}, so 2= \phi(2)=\phi(\sqrt{2})^2=a^2+3*b^2+2*a*b*\sqrt{3}. Since the \sqrt{3} is irrational we must have either a=0 or b=0. Then we use either \sqrt{2} or \sqrt{2/3} irrational to conclude it is impossible to solve this equation with rational numbers.
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  7. #7
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    Quote Originally Posted by canuhelp View Post
    hi thanks for your helps,i asked the same question under advanced topics
    and a friend give an answer what do you think about it
    http://www.mathhelpforum.com/math-he...-homework.html
    Quote Originally Posted by vemrygh View Post
    Construct a map \phi:\mathbb{Q}[\sqrt{2}]\rightarrow\mathbb{Q}[\sqrt{3}] such that: \phi(\sqrt{2})=\sqrt{3} and \phi(q)=q, for all q\in\mathbb{Q}.
    Then check that \phi is bijective and is a homomorphism, to show that \phi is an isomorphism.
    By the suggestion above:

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

    Thus the suggestion is not well defined!!

    canuhelp, did you read post #3?

    Also, I dont think its a good idea to post the same question in multiple places. It is against forum rules.
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  8. #8
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    dear Isomorphism canyou give us a complete explanation about this question as thinking this question to find a RİNG isomorphism beetween
    Q[\sqrt{2}]->Q[\sqrt{3}]
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  9. #9
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by vemrygh View Post
    Construct a map \phi:\mathbb{Q}[\sqrt{2}]\rightarrow\mathbb{Q}[\sqrt{3}] such that: \phi(\sqrt{2})=\sqrt{3} and \phi(q)=q, for all q\in\mathbb{Q}.
    Then check that \phi is bijective and is a homomorphism, to show that \phi is an isomorphism.

    That doesn't work as then 2 maps to 3 and so 1 maps to 3/2, which is absurd:

    \sqrt{3} = (\sqrt{2})\phi = (\sqrt{2}) \phi*(1) \phi = \sqrt{3}*3/2. Thus 3/2 = 1.
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by canuhelp View Post
    How can i prove that Q[_{\sqrt{2}}] is a ring isomorphism to Q[_{\sqrt{3}}]

    What does the question ask you? "Give an isomorphism..." or "Is there an isomorphism..."?

    I suspect it is the latter; the solution boils down to the question of "is \sqrt{3/2} a rational number", which i don't know off the top of my head.

    Hint: Assume there exists an isomorphism. Some element maps to \sqrt{3} - what must it look like?

    EDIT: I am entirely confused - I thought I was only the second person to post in this thread, after vemrygh. Thus my repetition of other peoples posts. I'm sure I read the entire thread first though!
    Last edited by Swlabr; May 14th 2009 at 08:17 AM.
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