# 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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This is just one of the Isomorphism Theorems. It holds for groups, rings, modules, and in general any algebra (in the sense of universal algebra). The proofs are all the same; in fact, you can take the proof for groups and it will be come a proof for rings mutatis mutandis. Here it is, explicitly, for rings.

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.

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By "in the other direction" do you mean that the map defined in the first part is surjective or just a Cantor-Bernstein kind of argument? – Asaf Karagila Oct 3 '11 at 17:12
I mean that the first map is surjective. – Henning Makholm Oct 3 '11 at 17:14