Help..

**kalagota** · December 1st 2007, 05:43 PM

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!!

**ThePerfectHacker** · December 1st 2007, 06:08 PM

Originally Posted by kalagota

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 $N=R$ .

**ThePerfectHacker** · December 1st 2007, 06:09 PM

Originally Posted by kalagota

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.

**kalagota** · December 1st 2007, 08:23 PM

Originally Posted by ThePerfectHacker

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..

**ThePerfectHacker** · December 2nd 2007, 06:53 AM

Originally Posted by kalagota

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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