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  1. #1
    MHF Contributor kalagota's Avatar
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    Help..

    1) Show that a factor ring of a field is either the trivial (zero) ring of one element or is isomorphic to the field.

    2) Let {\it R} and {\it R'} be rings and let {\it N} and {\it N'} be ideals of {\it R} and {\it R'}, respectively. Let {\it \phi} be a homomorphism of {\it R} into {\it R'}. Show that {\it \phi} induces a natural homomorphism {\phi_* : R/N \rightarrow R'/N'} if {\it \phi \left[ N \right] \subseteq N'}.

    Thanks!!
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  2. #2
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    Quote Originally Posted by kalagota View Post
    1) Show that a factor ring of a field is either the trivial (zero) ring of one element or is isomorphic to the field.
    Hint: Let N be an ideal of a unitary ring R. Show that if N contains a unit then N=R.
    Last edited by ThePerfectHacker; December 1st 2007 at 07:22 PM.
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    Quote Originally Posted by kalagota View Post
    2) Let {\it R} and {\it R'} be rings and let {\it N} and {\it N'} be ideals of {\it R} and {\it R'}, respectively. Let {\it \phi} be a homomorphism of {\it R} into {\it R'}. Show that {\it \phi} induces a natural homomorphism {\phi_* : R/N \rightarrow R'/N'} if {\it \phi \left[ N \right] \subseteq N'}.
    Any y\in R/N be be written as xN define \phi_*(xN) =\phi(x)N'. Now show it is well-defined.

    LaTeX tip: use \mapsto instead.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Any y\in R/N be be written as xN define \phi_*(xN) =\phi(x)N'. Now show it is well-defined.

    LaTeX tip: use \mapsto instead.
    so you mean, if y_1 , y_2 \in R/N such that y_1 = y_2, then i have to show that \phi_*(y_1) =\phi_*(y_2)

    ok.. if y_1 = x_1N and y_2 = x_2N for some x_1, x_2 \in R..

    \phi_*(y_1) = \phi(x_1)N' and \phi_*(y_2) = \phi(x_2)N'

    \implies \phi(x_1)N' = \phi(x_2)N' \Longleftrightarrow \phi(x_1)\phi(x_2)^{-1} = \phi(x_1)\phi(x_2^{-1}) = \phi(x_1x_2^{-1}) \in N'

    i this right??

    EDIT: scrap this.. i think, i got it already.. but if someone could write his proof, please do so, so that i can compare it with mine..
    Last edited by kalagota; December 2nd 2007 at 03:39 AM.
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  5. #5
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    Quote Originally Posted by kalagota View Post
    so you mean, if y_1 , y_2 \in R/N such that y_1 = y_2, then i have to show that \phi_*(y_1) =\phi_*(y_2)

    ok.. if y_1 = x_1N and y_2 = x_2N for some x_1, x_2 \in R..

    \phi_*(y_1) = \phi(x_1)N' and \phi_*(y_2) = \phi(x_2)N'

    \implies \phi(x_1)N' = \phi(x_2)N' \Longleftrightarrow \phi(x_1)\phi(x_2)^{-1} = \phi(x_1)\phi(x_2^{-1}) = \phi(x_1x_2^{-1}) \in N'

    i this right??
    Yes, and the last step is true because the image of N under phi is N'. That is how you use that little fact in the proof.

    Did you understand my other post?
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