Quotient rings and product of ideals question. Let $J \subset I$ ideals of a commutative ring R with multiplicative idetity $(1 \in R)$ . I dont know if the next fact is true or false:
If $a \in I^i$ but $a \not\in I^{i+1}$ for some integer $i\geq 1$, then $\overline{a} \in (I/J)^{i}$ but $\overline{a} \not\in (I/J)^{i+1}$. Where $\overline{a}$ is the class of $a$ in $I/J$.
 A: By definition, $(I/J)^n$ is the ideal of $R/J$ generated by all products $s_1 \cdots s_n$, where $s_i \in I/J$.  From the definition, it follows that $(I/J)^n = I^n/J$, where $I^n/J$ is the image of $I^n$ under the ring homomorphism $R \rightarrow R/J$.
We are given that $J$ is contained in $I$.  Let $n$ be the largest integer such that $J \subseteq I^n$ (possibly $n = \infty$).  Then your claim holds for all $i < n$: by the correspondence theorem, $\mathfrak a \mapsto \mathfrak a /J$ is an order preserving bijection from ideals of $R$ containing $J$ onto ideals of $R/J$.  It follows that if $x \in R$, then $x\in I^i$ if and only if $x + J \in  (I/J)^i$.  
Thus if $x \in I^i$, but not $I^{i+1}$, then $x+ J \in (I/J)^i$, but not in $(I/J)^{i+1}$.
On the other hand, if $J \not\subseteq I^{i+1}$, then it may be the case that $I^{i+1}$ is properly contained in $I^i$, but the images of $I^i$ and $I^{i+1}$ in $R/J$ are identical, making what you want impossible.  By the correspondence theorem for ideals, we can produce a counterexample to your claim if we produce a ring $R$ with $J \subseteq I$ ideals, and $I^{i+1} \subsetneq I^i$, and  $I^i + J = I^{i+1} + J$
Example: let $R = \mathbb{Z}, I = 3\mathbb{Z}, J = 45\mathbb{Z}$.  Then $J \subseteq I, I^3 \subsetneq I^2$, but $$I^2 + J = I^3 + J = 9\mathbb{Z}$$
Thus the images of $I^2$ and $I^3$ in $R/J = \mathbb{Z}/45\mathbb{Z}$ are the same.
