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Thread: Ring isomorphism

  1. #1
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    Ring isomorphism

    Hello I want to solve the following problem:

    Let R = Z( $\displaystyle sqrt2$ ) = { a+b$\displaystyle sqrt2$|a,b belong to Z.

    1. Prove that R is integral domain.

    2. If I = ($\displaystyle sqrt 2 $) is the ideal generated by $\displaystyle sqrt(2)$, prove that R/I~=Z2 (~= means isomorphic).


    //////

    In 1. I showed that elements of R are commutative and without zero-divisors.


    In 2. I understood I as all elements of R multiplied by $\displaystyle sqrt(2)$) . Am I correct?

    How shall I prove isomorphism? I tried the following function from R to Z2: f(a+$\displaystyle sqrt2$b+I)= a mod 2. But it is only hommomorphism in my opinion, not isomorphism.


    Can you help me?

    Thank you in advance!
    Last edited by andreas; Nov 16th 2008 at 04:38 AM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by andreas View Post
    Hello I want to solve the following problem:

    Let R = Z( $\displaystyle sqrt2$ ) = { a+b$\displaystyle sqrt2$|a,b belong to Z.

    2. If I = ($\displaystyle sqrt 2 $) is the ideal generated by $\displaystyle sqrt(2)$, prove that R/I~=Z2 (~= means isomorphic).

    In 2. I understood I as all elements of R multiplied by $\displaystyle sqrt(2)$) . Am I correct?

    How shall I prove isomorphism? I tried the following function from R to Z2: f(a+$\displaystyle sqrt2$b+I)= a mod 2. But it is only hommomorphism in my opinion, not isomorphism.
    The mapping $\displaystyle f:R\to\mathbb{Z}_2$ defined by $\displaystyle f(a+b\sqrt{2})=a\,({\rm mod}\,2)$ is an homomorphism, and it is obviously surjective. As a consequence, it suffices to show that its kernel is $\displaystyle I$ to conclude that $\displaystyle R/I\simeq \mathbb{Z}_2$.
    Its kernel consists of numbers $\displaystyle x=a+b\sqrt{2}$ where $\displaystyle a,b\in\mathbb{Z}$ and $\displaystyle a=2a'$ is even. Can you go on?
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  3. #3
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    Thank you! It was very helpful!! I can go on
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