# Bijection between ideals of $R/I$ and ideals containing $I$

I read that there is a one-one correspondence between the ideals of $R/I$ and the ideals containing $I$. ($R$ is a ring and $I$ is any ideal in $R$)

Is this bijection obvious? It's not to me. Can someone tell me what the bijection looks like explicitly? Many thanks for your help!

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Let $R$ be a ring, let $I$ be an ideal. The one-to-one correspondence between subrings of $R/I$ and subrings of $R$ that contains I$(which in fact also makes ideals correspond to ideals) is given as follows: Let$\pi\colon R\to R/I$be the canonical projection sending$r$to the class$r+I$. Given a subring$S$of$R$with$I\subseteq S\subseteq R$, we let $$\pi(S) = \{\pi(s)\mid s\in S\} = \{s+I\mid s\in S\}\subseteq R/I.$$ Given a subring$T$of$R/I$, we make it correspond to $$\pi^{-1}(T) = \{r\in R\mid \pi(r)\in T\}.$$ 1.$\pi(S)$is a subring of$R/I$whenever$S$is a subring of$R$that contains$R/I$. If$S$is a (left, right, two-sided) ideal, then$\pi(S)$is a (left, right, two-sided) ideal of$R/I$. Proof.$0\in S$, so$\pi(0) = 0+I \in \pi(S)$, hence$\pi(S)$is not empty. Also, if$(s+I),(t+I)\in \pi(S)$, with$s,t\in S$, then$s-t\in S$, so$(s+I)-(t+I) = (s-t)+I = \pi(s-t)\in \pi(S)$. Thus,$\pi(S)$is a subgroup of$R/I$. And if$s+I,t+I\in\pi(S)$, with$s,t\in S$, then$(s+I)(t+I) = st+I = \pi(st)\in \pi(S)$(since$S$is a subring of$R$), so$\pi(S)$is a subring. If in addition$S$is a (left) ideal of$R$, then given$(s+I)\in \pi(S)$and$(a+I)\in R/I$, with$s\in S$, we have$(a+I)(s+I) = as+I = \pi(as)$; since$S$is a (left) ideal,$s\in S$and$a\in R$, then$as\in S$, so$\pi(as)\in \pi(S)$. Similar arguments establish the right and two-sided cases. 2. If$T$is a subring of$R/I$, then$\pi^{-1}(T)$is a subring of$R$that contains$I$. If$T$is a (left, right, two-sided) ideal, then so is$\pi^{-1}(T)$. Proof.$0+I\in T$, and since for all$a\in I$,$\pi(a)=a+I = 0+I\in T$, then$a\in \pi^{-1}(T)$. Thus,$\pi^{-1}(T)$contains$I$. If$r,s\in \pi^{-1}(T)$, then so are$r-s$and$rs$, since$\pi(r-s) = (r-s)+I = (r+I)-(s+I)\in T$(since$r+I,s+I\in T$and$T$is a subring) and$\pi(rs) = rs+I = (r+I)(s+I)\in T$(since$T$is closed under products and$r+I,s+I\in T$). Thus,$\pi^{-1}(T)$is a subring of$R$. If$T$is a left ideal of$R/I$, and$s\in\pi^{-1}(T)$,$a\in R$, then$\pi(s)\in T$, so$\pi(as) = \pi(a)\pi(s)\in T$(since$T$is a left ideal), so$as\in\pi^{-1}(T)$. Thus,$\pi^{-1}(T)$is a left ideal of$R$. Similar arguments establish the right and two-sided cases. 3. The correspondences are inverses of each other, hence they are bijections. Proof. Let$S$be an ideal of$R$that contains$I$. Then$S\subseteq \pi^{-1}(\pi(S))$holds, because it holds for any subset and any function. Now let$a\in \pi^{-1}(\pi(S))$. then$\pi(a)\in \pi(S)$, so there exists$s\in S$such that$\pi(a)=\pi(s)$; hence$\pi(a-s)\in\mathrm{ker}(\pi) = I$. Thus,$a-s\in I\subseteq S$. Since$a-s,s\in S$, and$S$is a subring of$R$, then$a=(a-s)+s\in S$. Thus,$\pi^{-1}(\pi(S))\subseteq S$, proving equality. Conversely, if$T$is an ideal of$R/I$, then$\pi(\pi^{-1}(T))=T$, because$\pi$is onto and this equality holds for any surjective function.$\Box$4. The correspondences are inclusion-preserving. Proof. For any function$f\colon X\to Y$and subsets$A,B\subseteq X$, if$A\subseteq B$then$f(A)\subseteq f(B)$; and for any subsets$C,D$of$Y$, if$C\subseteq D$then$f^{-1}(C)\subseteq f^{-1}(D)$, so this follows from purely set-theoretic considerations. - Let$J\supseteq I$be an ideal of$R$. Because$I$is closed under negation and$J$is closed under addition, each coset of$I$is either contained in$J$or disjoint from$J$, and thus$J$maps directly to a subset of$R/I$via the canonical projection homomorphism$\pi:R\to R/I$; the image happens to be an ideal. In the other direction, assume$K$is an ideal in$R/I$. Then$\pi^{-1}(K)$is easily seen to be an ideal of$R$(the preimage of an ideal under a surjective homomorphism is always an ideal); it contains$I$because$0_{R/I}\$ must be in every ideal.