# How is this map a well-defined homomorphism?

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.

-

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:

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

-
Thank you and @complex analysis. – PandaMan Dec 3 '13 at 21:44
You say $r+I$ belongs to $I$??? I think $r+I$belongs to $R/I$. – Sara Apr 17 '15 at 6:27
@sara I can't see where you are getting that from. Can you be specific about what you are reading? – rschwieb Apr 17 '15 at 9:57

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.

-
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