Having trouble with this following question:
Let $R$ be a ring and $I$ and $J$ ideals of $R$ such that $J\subset I$ ($J\subseteq I, J\neq I $)
a) Show that the set $I/J$ is ideal of quotient ring $R/J$
b) Show that the mapping $f: R/J \rightarrow R/I$, $f(r+J)=r+I$ is well defined.
c) Show that $(R/J)/(I/J) \cong R/I $
My attempt:
a) $I/J=\{a+J | a\in I\}$ ideal of $R/J$?
- $I/J \neq \emptyset$, since $J\subset I$ and $I \neq J$
- Let $a,b\in I/J$. Now $a - b =(a'+J)-(b'+J)=a'-b'+J \in I/J$, since $I$ is ideal of $R$ and therefore $a'-b'$ must be in $I$, and $a'-b'+J \in I/J$.
- Let $a\in I/J$ and $r\in R/J$.
$ar = (a'+J)(r'+J)=a'r'+J\in I/J$ with the same reasoning as recently.
Applying the same for case $ra$.
Is this proof rigorous enough?
b) Don't really know how to proceed on part b. I think it should be shown that mapping is independent of representative of side class? But not sure how to approach it.
c) I noticed that $(R/J)/(I/J) =\{r+I|r\in R\}=R/I$, so $\#(R/J)/(I/J)=\#R/I$. Then it should be used some mapping (maybe use part b?) to show the bijectiveness and isomporphism of the rings.
Any help will be appreciated.
a) It's correct, but I don't see where is it that you think you used the fact that $J\neq I$.
b) Suppose $r+J=r'+J$. This means that $r-r'\in J$. But since $J\subset I$, then $r-r'\in I$. Therefore, $r+I=r'+I$ and so the map is well-defined.
c) $\ker f=I/J$ and $f$ is surjective. So, since $\operatorname{Im}(f)\simeq(R/I)/\ker f$, you have $(R/I)/(I/J)\simeq R/J$.