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

    Thanks!!
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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 $\displaystyle N$ be an ideal of a unitary ring $\displaystyle R$. Show that if $\displaystyle N$ contains a unit then $\displaystyle N=R$.
    Last edited by ThePerfectHacker; Dec 1st 2007 at 06:22 PM.
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    Quote Originally Posted by kalagota View Post
    2) Let $\displaystyle {\it R}$ and $\displaystyle {\it R'}$ be rings and let $\displaystyle {\it N}$ and $\displaystyle {\it N'}$ be ideals of $\displaystyle {\it R}$ and $\displaystyle {\it R'}, $respectively. Let $\displaystyle {\it \phi}$ be a homomorphism of $\displaystyle {\it R}$ into $\displaystyle {\it R'}. $Show that $\displaystyle {\it \phi}$ induces a natural homomorphism $\displaystyle {\phi_* : R/N \rightarrow R'/N'}$ if $\displaystyle {\it \phi \left[ N \right] \subseteq N'}$.
    Any $\displaystyle y\in R/N$ be be written as $\displaystyle xN$ define $\displaystyle \phi_*(xN) =\phi(x)N'$. 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 $\displaystyle y\in R/N$ be be written as $\displaystyle xN$ define $\displaystyle \phi_*(xN) =\phi(x)N'$. Now show it is well-defined.

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

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

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

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

    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; Dec 2nd 2007 at 02:39 AM.
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    Quote Originally Posted by kalagota View Post
    so you mean, if $\displaystyle y_1 , y_2 \in R/N$ such that $\displaystyle y_1 = y_2$, then i have to show that $\displaystyle \phi_*(y_1) =\phi_*(y_2)$

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

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

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

    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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