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Thread: Ring Isomorphism

  1. #1
    Super Member Bernhard's Avatar
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    Ring Isomorphism

    Can anyone help with the following problem:

    Prove that the rings $\displaystyle 2\mathbb{Z}$ and $\displaystyle 3\mathbb{Z}$ are not isomorphic?

    Be grateful for help

    Peter
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  2. #2
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    Re: Ring Isomorphism

    How many members does $\displaystyle 2\mathbb{Z}$? What about $\displaystyle 3\mathbb{Z}$?
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Ring Isomorphism

    2Z = { ... -4.-2, 0, 2, 4, 6, ....} so it is infinite

    3Z = { ... -6. -3, 0, 3, 6, .... } so it is infinite

    ... or have I got something wrong?

    Peter
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  4. #4
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    Re: Ring Isomorphism

    Quote Originally Posted by Bernhard View Post
    Can anyone help with the following problem:

    Prove that the rings $\displaystyle 2\mathbb{Z}$ and $\displaystyle 3\mathbb{Z}$ are not isomorphic?

    Be grateful for help

    Peter

    Let $\displaystyle f:2\mathbb{Z} \to 3\mathbb{Z}$ be a ring isomorphism. Then $\displaystyle f(2) \in 3\mathbb{Z}$. Thus $\displaystyle f(2) = 3 x$, $\displaystyle x \in \mathbb{Z}$.

    Therefore $\displaystyle f(4) = f(2+2) = f(2)+f(2) = 6x$ and $\displaystyle f(4)=f(2\cdot2)=f(2) f(2)=9x^2$.

    So, $\displaystyle 6x=9x^2 \Rightarrow x=0$ or $\displaystyle x=2/3$.

    $\displaystyle x=0$ means $\displaystyle f(2)=0$, so $\displaystyle f(2\lambda ) = f(2)f(\lambda )=0$, $\displaystyle \forall \lambda \in \mathbb{Z}$ which means that $\displaystyle f $ is not an isomorphism,

    and $\displaystyle x=2/3$ with $\displaystyle x\in \mathbb{Z}$ is Absurd.

    So, the rings $\displaystyle 2\mathbb{Z}$ and $\displaystyle 3\mathbb{Z}$ are not isomorphic.
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  5. #5
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    Re: Ring Isomorphism

    Sorry, I was thinking "$\displaystyle \mathbb{Z}$ modulo 2 and 3". zoek has it right.
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  6. #6
    Super Member Bernhard's Avatar
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    Re: Ring Isomorphism

    No problems. Thanks anyway

    And thanks for help with previous posts as well!

    Peter
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