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Math Help - isomorphism

  1. #1
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    isomorphism

    Let y be an integer, which is not a square of another integer.

    How do I prove that Q[sqrt(y)] = {a + bsqrt(y) : a,b in Q} is isomorphic to Q[x]/(x^2 - y) where Q is the set of rational numbers?

    Thanks for your help.
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  2. #2
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    Quote Originally Posted by agentZERO View Post
    Let y be an integer, which is not a square of another integer.

    How do I prove that Q[sqrt(y)] = {a + bsqrt(y) : a,b in Q} is isomorphic to Q[x]/(x^2 - y) where Q is the set of rational numbers?

    Thanks for your help.
    I think someone is replying to this now who should be able to help, so since I'm kind of learning this topic as well, I'll ask a question.

    This proof requires that we show that there exists a map F:Q[\sqrt{y} ]\rightarrow \frac{Q[x]}{x^2-y} that is one-to-one, onto, and has an inverse. That is the more explicit objective right?

    What is x defined as though?
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  3. #3
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    Quote Originally Posted by agentZERO View Post
    Let y be an integer, which is not a square of another integer.

    How do I prove that Q[sqrt(y)] = {a + bsqrt(y) : a,b in Q} is isomorphic to Q[x]/(x^2 - y) where Q is the set of rational numbers?

    Thanks for your help.
    define the map f: \mathbb{Q}[x] \longrightarrow \mathbb{Q}[y] by f(p(x))=p(\sqrt{y}). clearly f is an onto ring homomorphism. so we only need to prove that \ker f = <x^2 - y>, which is quite straightforward:

    if p(x) \in <x^2 - y>, then obviously p(\sqrt{y})=0, i.e. p(x) \in \ker f. suppose now that p(x) \in \ker f. we have: p(x)=(x^2 -y)q(x) + \alpha x + \beta, for some \alpha, \beta \in \mathbb{Q}. since p(\sqrt{y})=0,

    we must have: \alpha \sqrt{y} + \beta = 0, which gives us: \alpha = \beta = 0, because y is not a square of an integer. thus: p(x)=(x^2 - y)q(x) \in <x^2 -y>. \ \ \Box
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  4. #4
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    Oh man was I wrong. Sigh. So much to learn.
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  5. #5
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    How do I use this to prove that Q[sqrt(2)] is not isomorphic to Q[sqrt(3)]? What property of a ring fails?

    Thanks for the response from before.
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  6. #6
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    Quote Originally Posted by agentZERO View Post
    How do I use this to prove that Q[sqrt(2)] is not isomorphic to Q[sqrt(3)]? What property of a ring fails?

    Thanks for the response from before.
    This is an interesting problem that \mathbb{Q}(\sqrt{2}) and \mathbb{Q}(\sqrt{3}) are isomorphic as vector spaces but not as fields. The way you show they are not isomorphic is by assuming that \phi: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3}) is an isomorphism. First \phi ( 1 ) = \phi (1). Therefore, \phi (n ) = \phi (1 + ... + 1) = \phi (1 ) + ... + \phi (1 ) = n for n\in \mathbb{Z}^+. It is easy to show \phi ( n ) = n for n\in \mathbb{Z} by considering the negatives. Next \phi(a/b) = \phi(a)/\phi(b) = a/b. Thus, \phi|_{\mathbb{Q}} is an identity map. Now \sqrt{2}^2 = 2 \implies \phi(\sqrt{2})^2 = 2 \implies \phi(\sqrt{2}) = \pm \sqrt{2}. In both cases we never get an onto mapping because nothing gets mapped to \sqrt{3}. A contradiction.
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