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Math Help - 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 \phi? is it like this?

    let F , F' be fields and define \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 \phi? is it like this?

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

    thanks..
    It is hard to understand your difficulty.
    If \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 \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 \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 \phi?
    ok, check out my "proof"..

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

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

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

    therefore \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 \phi is a homomrophism from the field then \ker \phi is an ideal. But a field can only has \{ 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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