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Math Help - Is my proof correct? (Isomorphism)

  1. #1
    Senior Member sfspitfire23's Avatar
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    Is my proof correct? (Isomorphism)

    We must show that \mathbb{R} and \mathbb{C} are NOT isomorphic. Consider the mapping \alpha : \mathbb{C} \rightarrow \mathbb{R}. Then we have \alpha(i)=a which can be written as  \alpha(i)^2=a^2. So,  \alpha(i^2)=a^2 which is the same as \alpha(-1)=a^2. We can split this apart to get \alpha(-1)+\alpha(0)=a^2. This shows that -1=a^2. But nothing in \mathbb{R} squares to  -1 . Therefore they are not isomophic.
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  2. #2
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    \alpha(-1)=-1 since \alpha is an homomorphism, I don't know why you split this in the last term. But, yeah your proof is fine.
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    Quote Originally Posted by sfspitfire23 View Post
    We must show that \mathbb{R} and \mathbb{C} are NOT isomorphic. Consider the mapping \alpha : \mathbb{C} \rightarrow \mathbb{R}. Then we have \alpha(i)=a which can be written as  \alpha(i)^2=a^2. So,  \alpha(i^2)=a^2 which is the same as \alpha(-1)=a^2. We can split this apart to get \alpha(-1)+\alpha(0)=a^2. This shows that -1=a^2. But nothing in \mathbb{R} squares to  -1 . Therefore they are not isomophic.

    Almost: why \alpha(-1)=-1 ?

    Tonio
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  4. #4
    Senior Member sfspitfire23's Avatar
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    Quote Originally Posted by tonio View Post
    Almost: why \alpha(-1)=-1 ?

    Tonio


    Because it's a homomorphism? I think this is where I slip up. I'm not sure how to say that it goes to -1.
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    Quote Originally Posted by sfspitfire23 View Post
    Because it's a homomorphism? I think this is where I slip up. I'm not sure how to say that it goes to -1.

    I supose you were trying to show \mathbb{C}\ncong \mathbb{R} as rings, and then \alpha(1)=1\,,\,\alpha(0)=0\,\,\Longrightarrow\,\, 0=\alpha(0)=\alpha(1+(-1))=\alpha(1)+\alpha(-1)=1+\alpha(-1) and we're done.

    Tonio
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