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If $f: R \rightarrow S$ is a homomorphism of rings with kernel $K$, and $I$ is an ideal in $R$ such that $I \subset K$.

The hypothesis is that the map $\overline{f}: R/I \rightarrow S$ given by $\overline{f}(r+I)=f(r)$ is a well-defined homomorphism.

For the well-defined part, I know that I need to somehow show that when $a=b$ in $R/I$, then $\overline{f}(a)=\overline{f}(b)$ in $S$. I just don't know where to start.

For the homomorphism part, $\overline{f}(ab)=\overline{f}(a)\overline{f}(b)$, I'm also unclear as how to proceed.

Thanks in advance!

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2 Answers 2

up vote 1 down vote accepted

Let's write out what it means for $r+I=s+I$, and then show that finally $f(r)=f(s)$.

We know that means $r-s\in I$. Since $r-s\in I\subseteq ker(f)$, this would imply that $f(r-s)=0$, and hence that $f(r)=f(s)$. Thus $r+I=s+I$ implies $f(r)=f(s)$, and thus the map is well defined.

As for the other computation:

$\overline{f}((r+I)(s+I))=\overline{f}(rs+I)=f(rs)=f(r)f(s)=\overline{f}(r+I)\overline{f}(s+I)$

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Thank you and @complex analysis. –  PandaMan Dec 3 '13 at 21:44

If , $a=b$ then its clear that $a+I=b+I \implies \bar f(a)= \bar f(b)$ . Sorry , after @rshweib made a important remark , the implication follows because $a+I=b+I$ means $a-b \in I$ ie . $f(a-b)=0$ hence $f(a)=f(b)$ For the second part $$\bar f (ab) = f(ab+I) = f((a+I)(b+I))= f(a+I)\cdot f(b+I) = \bar f(a)\bar f(b)$$ last equality follows from the fact the $f$ is a homomorphism.

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thanks! god bless. –  PandaMan Dec 3 '13 at 21:23
    
@rschwieb : right , i needed to argue with the fact that $a-b$ belongs to the kernel of $f$. sorry for leaving the important part . –  Complex analysis Dec 3 '13 at 21:27
    
@rschwieb : Is it ok now ? –  Complex analysis Dec 3 '13 at 21:31
    
@Complexanalysis Better, but there is still a notational problem. It looks like you're treating $\overline{f}$ as a function from $R$ and $f$ as a function from $R/I$, but that's not the case. They need to be switched. –  rschwieb Dec 4 '13 at 13:39

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