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Thread: help again

  1. #1
    MHF Contributor kalagota's Avatar
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    help again

    Show that each homomorphism of a field is either one to one or maps everything to 0..

    the one to one part is easy, i need the second one..
    also, how will i define $\displaystyle \phi$? is it like this?

    let F , F' be fields and define $\displaystyle \phi : F \rightarrow F'$... i am not sure..

    thanks..
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  2. #2
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    Quote Originally Posted by kalagota View Post
    Show that each homomorphism of a field is either one to one or maps everything to 0.. The one to one part is easy, i need the second one.. also, how will i define $\displaystyle \phi$? is it like this?

    let F , F' be fields and define $\displaystyle \phi : F \rightarrow F'$... i am not sure..

    thanks..
    It is hard to understand your difficulty.
    If $\displaystyle \phi$ maps every element to zero, then you are done.
    Otherwise, show it is injective. You said that is the easy part.
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Plato View Post
    It is hard to understand your difficulty.
    If $\displaystyle \phi$ maps every element to zero, then you are done.
    Otherwise, show it is injective. You said that is the easy part.
    thanks, but how will i define phi?
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  4. #4
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    Quote Originally Posted by kalagota View Post
    Show that each homomorphism of a field is either one to one or maps everything to 0..
    I don't see where we are ask to define $\displaystyle \phi$?
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Plato View Post
    I don't see where we are ask to define $\displaystyle \phi$?
    ok, check out my "proof"..

    let $\displaystyle a,b \in F$ such that $\displaystyle \phi (a) = \phi (b)$

    then $\displaystyle \phi (a) - \phi (b) = \phi (a-b) = 0$

    but by properties of a homomorphism, $\displaystyle a - b = 0 \implies a=b$

    therefore $\displaystyle \phi$ is 1-1.. QED.

    is that it?
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  6. #6
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    Quote Originally Posted by kalagota View Post
    Show that each homomorphism of a field is either one to one or maps everything to 0..
    Theorem: The kernel of a ring homomorphism is an ideal of ring.

    Thus, if $\displaystyle \phi$ is a homomrophism from the field then $\displaystyle \ker \phi$ is an ideal. But a field can only has $\displaystyle \{ 0\}$ or itself as an ideal. Thus, in the first case the map is one-to-one because the kernel is trivial. And in the other case the map collapses everything into 0 because it is the full field.
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