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Math Help - Question about Ring Homomorphism

  1. #1
    Senior Member slevvio's Avatar
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    Question about Ring Homomorphism

    Hello I just had a quick question about ring homomorphisms.

    Is it possible to have a ring isomorphism from R to F, where F is a field but R is not a field?
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  2. #2
    Senior Member abhishekkgp's Avatar
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    edit: deleted post.

    hi slevvio!!

    f: R----> F
    f(x)=0 for all x in R.

    this is trivial homomorphism i guess. Is this acceptable to you??
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  3. #3
    Senior Member slevvio's Avatar
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    Thanks but that is not a ring isomorphism!
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  4. #4
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    if φ is onto, R has to be a field, since then R = φ^-1(F).

    if φ is not onto, and φ(R) is just a subring of F, then sure, consider the identity homomorphism which includes Z in Q.
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  5. #5
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by slevvio View Post
    Thanks but that is not a ring isomorphism!
    my book says that if R and S are rings then f:R ---> S is a homomorphism if the following hold for all a,b in R

    1) f(a+b)=f(a)+f(b)
    2)f(ab)=f(a)f(b)

    aren't these two satisfied in the example i gave??
    am i missing something out?
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  6. #6
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    Quote Originally Posted by abhishekkgp View Post
    my book says that if R and S are rings then f:R ---> S is a homomorphism if the following hold for all a,b in R

    1) f(a+b)=f(a)+f(b)
    2)f(ab)=f(a)f(b)

    aren't these two satisfied in the example i gave??

    am i missing something out?
    unless R is trivial, that is not an isomorphism.
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  7. #7
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Deveno View Post
    unless R is trivial, that is not an isomorphism.
    my bad! i misread the question.I'm sorry slevvio and denevo.
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  8. #8
    Senior Member slevvio's Avatar
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    Thanks guys. I was just confused because in my notes a ring homomorphism was defined from a ring to a field, and then after this was proved to be a ring isomorphism, he showed that the initial ring was actually a field. But this should arise from the fact we have a ring isomorphism. I would type out the specific example but Latex is borked
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