I know the elements of $\frac{R}{J}$ are like $r+J$ when $r \in R$. So I have to show that there is some $s \in R$ that $r+J = s+I$. But I don't know how to do that. I actually want to show that $\vert{\frac{R}{J}}\vert \lt \vert{\frac{R}{I}}\vert$ and I thought I can prove it by saying that $\frac{R}{J} \subset \frac{R}{I}$. Am I doing right? If not, How can I prove $\vert{\frac{R}{J}}\vert \lt \vert{\frac{R}{I}}\vert$?
Would you please explain it to me?
On
First, on a the level of equality what you are asking for isn't generally going to happen. To see why, let's look at an example. Let $R= \mathbb Z$, $I = 4\mathbb Z$, and $J = 2\mathbb Z$. Let's look at $1 + J =1 +\{\dots-4, -2 , 0 ,2, 4, \dots\} = \{\dots, -1, 1, 3, \dots\}$. Hypothetically, we to find an $s$ such that $1+J = s+ I = s + \{\dots,-8,-4,0,4,8,\dots\} = \{\dots, -4+s, s, 4+s, \dots\}$. But this is never going to happen as a matter of pure equality. The thing is that the elements are different kinds of objects. Instead, what you should be asking for is some way to embed $R/J$ in $R/I$ (ie: an injective ring homomorphism from $R/J$ to $R/I$.
However, I'm pretty sure that you aren't going to be able to do that in full generality with an explicit construction. My intuition comes from the fact that purely on a set-theoretic level, we can't prove $|R/I| \leq |R|$ in full generality without appealing to the axiom of choice. Namely, we can't in general explicitly construct an injective map from $R/I$ to $R$. However, if we do assume choice, then we know that if you have a surjection $R\to R/I$ we have an injection $R/I \to R$. There might be some subtlety due to the fact that these are rings which makes proving this possible where it might not be possible for general sets, but I doubt it.
So what are we to do. Well, we probably aren't going to generally be able to construct an injection, but when a door closes a window opens. If we are assuming choice, then it is enough to prove that there is a surjective map from $R/I$ to $R/J$. Do you have any ideas about how to do that?
In general is not true that for every $r$ there is $s$ such that $$r+J=s+I$$ for example, take $r\in J$, then the equation became $J=s+I$, which have no solution unless $s\in I$ (because either $s\in I$ or $s+I$ is not an ideal, in particular it cannot be $=J$), so you would get $I=J$.
If you have two ideals $I\subseteq J$, then there is a canonical map $\pi:R/I\rightarrow R/J$ mapping $x+I$ to $x+J$. This map is onto and it's kernel is $J/I$ (simply check the definition, $r+I\in\ker \pi$ iff $r+J=J$, that is, if $r\in J$, so $\ker\pi=J/I$), therefore by the first isomorphism theorem $$\frac{R/I}{J/I}\simeq R/J$$ so $$|R/J|=\left|\frac{R/I}{J/I}\right|\leqslant R/I$$ where in the last inequality I used the axiom of choice: The projection $\pi:R/I\rightarrow (R/I)/(J/I)$ is onto, therefore it has a section $\sigma:(R/I)/(J/I)\rightarrow R/I$ (which, in general, is just a function, and not a ring homomorphism) which is 1-1. The reason I've used $\leqslant$ instead of $<$ is because when the quotients are infinite, they might have same cardinality (even if one ideal is strictly contained in the other), for example, take $R=\mathbb R$, $I=\mathbb Z$ and $J=\mathbb Z[\frac{1}{2}]=\{n+\frac{m}{2}|n,m\in\mathbb Z\}$, then $$R/I\simeq R/J\simeq\mathbb S^1=\{z\in\mathbb C:|z|=1\}$$