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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!

Rudy the Reindeer
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2 Answers2

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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 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.

exchange
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Arturo Magidin
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    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 $R/I$."? – exchange Jul 06 '18 at 09:59
  • @exchange: yes; feel free to make the correction. – Arturo Magidin Jul 06 '18 at 17:20
  • Is there an example of mapping that's not inclusion-preserving ? – omg Sep 26 '21 at 00:52
  • @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?" – Arturo Magidin Sep 26 '21 at 01:02
  • 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 '21 at 01:53
  • @omg: What does "always be the case" even mean? In what circumstances? Note that the function is not "inclusion preserving" on all ideals of $R$, just on some. And not every function is inclusion preserving. It is mentioned because (i) it is true; (ii) it is not automatic; and (iii) it is important. – Arturo Magidin Sep 26 '21 at 02:04
  • Because you mentioned that this follows from purely set-theoretic considerations., so I thought it's always the case(always establishes, sorry for poor english), isn't it? – omg Sep 26 '21 at 02:07
  • @omg Why are you using text quoting? Don't use a single apostrophe, it kills the formatting. Use quotation marks if you are trying to quote. The correspondence is inclusion preserving among some of the ideals because it is defined in terms of the set theoretic function. But you have to restrict which ideals you are considering. Not every correspondence is defined in terms of the direct image functor of a set theoretic function. Again, you mention it because it's true, it's important, it's not immediate. The latter is not always the case for an arbitrary correspondence between sets. – Arturo Magidin Sep 26 '21 at 02:11
  • Thanks a lot for clarification! – omg Sep 26 '21 at 02:14
  • In point 1 the hypothesis states, among other things, that "S contains I". But is this used in this part of the proof? It seems unneccessary. Am I missing something? – BodyDouble Aug 23 '23 at 21:25
  • @BodyDouble: Technically, you do not need it; the image of $S$ equals the image of $S+I$; but because what we are trying to establish is a bijection between the subrings of $R$ that contain $I$ and the subrings of $R/I$, we want to define a correspondence between these two sets, and verify that the correspondence is well-defined. While parts of the argument may apply to a larger collection of objects, there is no reason to state, in a proof, a more general argument than the one you need for the proof. – Arturo Magidin Aug 23 '23 at 21:37
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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.

hmakholm left over Monica
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