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Thread: 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 $\displaystyle F:Q[\sqrt{y} ]\rightarrow \frac{Q[x]}{x^2-y}$ 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 $\displaystyle f: \mathbb{Q}[x] \longrightarrow \mathbb{Q}[y] $ by $\displaystyle f(p(x))=p(\sqrt{y}).$ clearly $\displaystyle f$ is an onto ring homomorphism. so we only need to prove that $\displaystyle \ker f = <x^2 - y>,$ which is quite straightforward:

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

    we must have: $\displaystyle \alpha \sqrt{y} + \beta = 0,$ which gives us: $\displaystyle \alpha = \beta = 0,$ because $\displaystyle y$ is not a square of an integer. thus: $\displaystyle p(x)=(x^2 - y)q(x) \in <x^2 -y>. \ \ \Box$
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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 $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \mathbb{Q}(\sqrt{3})$ are isomorphic as vector spaces but not as fields. The way you show they are not isomorphic is by assuming that $\displaystyle \phi: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$ is an isomorphism. First $\displaystyle \phi ( 1 ) = \phi (1)$. Therefore, $\displaystyle \phi (n ) = \phi (1 + ... + 1) = \phi (1 ) + ... + \phi (1 ) = n$ for $\displaystyle n\in \mathbb{Z}^+$. It is easy to show $\displaystyle \phi ( n ) = n$ for $\displaystyle n\in \mathbb{Z}$ by considering the negatives. Next $\displaystyle \phi(a/b) = \phi(a)/\phi(b) = a/b$. Thus, $\displaystyle \phi|_{\mathbb{Q}}$ is an identity map. Now $\displaystyle \sqrt{2}^2 = 2 \implies \phi(\sqrt{2})^2 = 2 \implies \phi(\sqrt{2}) = \pm \sqrt{2}$. In both cases we never get an onto mapping because nothing gets mapped to $\displaystyle \sqrt{3}$. A contradiction.
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