# 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!

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 become 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 contain $I$ (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 $I$. 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. If $(\pi^{-1}\circ\pi)$ and $(\pi\circ\pi^{-1})$ are both the identity, then $\pi$ is an isomorphism.

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 $(\pi^{-1}\circ\pi)=\text{id}$.

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.

• Great answer. In 1., should it be "$\pi(S)$ is a subring of $R/I$ whenever $S$ is a subring of $R$ that contains $I$." instead of "... that contains $R/I$."? Jul 6, 2018 at 9:59
• @exchange: yes; feel free to make the correction. Jul 6, 2018 at 17:20
• Is there an example of mapping that's not inclusion-preserving ?
– omg
Sep 26, 2021 at 0:52
• @omg: You should have waited a week: it could have been exactly ten years after the answer was written... You can define maps any way you want. You can define them so they do not preserve anything. But what's the point? You want maps that are useful, or natural. If all you are asking is"can you define functions that don't preserve inclusions?" the answer is "yes. So what?" Sep 26, 2021 at 1:02
• My point is that inclusion-preserving seems to be always the case, so why mention it? BTW, thanks a lot for the quick answer!
– omg
Sep 26, 2021 at 1:53

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.

• 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? Oct 3, 2011 at 17:12
• I mean that the first map is surjective. Oct 3, 2011 at 17:14
• What does "closed under negation" here mean? What are negating? Dec 21, 2018 at 12:57
• @Hawk: $I$ is closed under negation means that for every $x\in I$ we also have $(-x) \in I$. Dec 21, 2018 at 16:28