Does this statement always true?(quotient ring)

Question

Statement

Let the ring $R$ and its ideal $I$ and $J$ s.t. $I\subset J$

Then there is a subring(or ideal) $R_J(\simeq R/J)$ of the $R/I$

There are 2 questions.

First) I'm not sure the above statement is right. If the above statement is correct, How could I prove it?

Second) Let's substitute as the normal groups $N_1$, $N_2$ and $G$(group) instead of the ideals $I$,$J$ and $R$(ring).

Then does it true when we considering the group case?

See this question https://math.stackexchange.com/questions/813978/correspondence-theorem-for-rings. This is a ring version of the correspondence theorem. — CardioidAss22, Jul 09 '19 at 11:33

score 1 · Accepted Answer · answered Jul 09 '19 at 12:39

First)

Not correct.

Surjective ringhomomorphism $\nu:R/I\to R/J$ prescribed by $r+I\mapsto r+J$ has kernel $J/I:=\{j+I\mid j\in J\}\subseteq R/I$.

Then $R/J$ and $(R/I)/(J/I)$ are isomorpic but $(R/I)/(J/I)$ is a quotient of $R/I$ is (not a subring or an ideal).

Second)

Yes, there is an analogy. Again we are dealing with a quotient then and not a subgroup.

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