1. ## 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!!

2. 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 $N=R$.

3. 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.

4. 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.

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..

5. 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?