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Math Help - Ring Isomorphism

  1. #1
    Junior Member hercules's Avatar
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    Ring Isomorphism - help needed

    I need help with the following proof:

    Show that if  \theta:Z \rightarrow Z is a ring isomorphism, then \theta must be the identity mapping. Is there an additive group isomorphism \theta:Z \rightarrow Z other than the identity mapping?

    Thanks

    and how do you Z that represents the set of integers in Latex.
    Last edited by hercules; April 26th 2008 at 05:43 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by hercules View Post
    I need help with the following proof:

    Show that if  \theta:Z \rightarrow Z is a ring isomorphism, then \theta must be the identity mapping. Is there an additive group isomorphism \theta:Z \rightarrow Z other than the identity mapping?

    Thanks

    and how do you Z that represents the set of integers in Latex.
    first \mathbb{Z} will give the integers symbol.

    for the 2nd part consider group \mathbb{Z},+

    then \theta(x)=-x is an isomorphism

    \theta(a+b)=-(a+b)=(-a)+(-b)=\theta(a)+\theta(b)
    and

    \theta (0)=-0=0

    For the first part i am not sure. I will think about it.

    Good luck.
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  3. #3
    Junior Member hercules's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    first \mathbb{Z} will give the integers symbol.

    for the 2nd part consider group \mathbb{Z},+

    then \theta(x)=-x is an isomorphism

    \theta(a+b)=-(a+b)=(-a)+(-b)=\theta(a)+\theta(b)
    and

    \theta (0)=-0=0

    For the first part i am not sure. I will think about it.

    Good luck.


    Thanks Emptyset.

    Can someone please help me with the first part. My major problem.
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  4. #4
    Behold, the power of SARDINES!
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    Maybe this is it?

    Quote Originally Posted by hercules View Post
    I need help with the following proof:

    Show that if  \theta:Z \rightarrow Z is a ring isomorphism, then \theta must be the identity mapping. Is there an additive group isomorphism \theta:Z \rightarrow Z other than the identity mapping?

    Thanks

    and how do you Z that represents the set of integers in Latex.
    I have been thinking about this and this is what i came up with...

    proof by contradiction

    let \Phi (x) an isomorphsim that is not the identity mapping.

    Then there exists a \in \mathbb{Z} such that

    a \ne \Phi (a)

    I will assume a is positive(it can be done if a is negative)

    we can write a in a few different ways

    a = 1 \cdot a=\underbrace{1+1+1+1+...+1}_{a Times}

    since Phi is an isomorphism we know that

    \Phi (0)=0,\Phi(1)=1

    so know we get

    \Phi(a)=\Phi(1+1+1+...+1)= \Phi(1)+\Phi(1)+ \Phi(1)+ \Phi(1)+ ...\Phi(1)=a \cdot \Phi(1) =a \cdot 1=a
    \Phi(a)=\Phi(1\cdot a)=\Phi(1) \cdot \Phi(a)=1\cdot \Phi(a)=\Phi(a)

    This is a contradiction.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I have been thinking about this and this is what i came up with...

    proof by contradiction

    let \Phi (x) an isomorphsim that is not the identity mapping.

    Then there exists a \in \mathbb{Z} such that

    a \ne \Phi (a)

    I will assume a is positive(it can be done if a is negative)

    we can write a in a few different ways

    a = 1 \cdot a=\underbrace{1+1+1+1+...+1}_{a Times}

    since Phi is an isomorphism we know that

    \Phi (0)=0,\Phi(1)=1

    so know we get

    \Phi(a)=\Phi(1+1+1+...+1)= \Phi(1)+\Phi(1)+ \Phi(1)+ \Phi(1)+ ...\Phi(1)=a \cdot \Phi(1) =a \cdot 1=a
    \Phi(a)=\Phi(1\cdot a)=\Phi(1) \cdot \Phi(a)=1\cdot \Phi(a)=\Phi(a)

    This is a contradiction.
    I was thinking the same thing. I don't know if that makes you more or less confident in your solution...
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  6. #6
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    If \theta: \mathbb{Z}\mapsto \mathbb{Z} is a ring isomorphisms then certainly it is a group automorphism of \left< \mathbb{Z},+\right>. Group isomorphism on cyclic groups are completely determined on the generator. Since we must have \theta (1) = 1 (for that is the definition of a commutative unitary ring homomorphism) it is completely determined, i.e. unique, and in fact the identity mapping.
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