Suppose that $(R,+,.)$ is a commutative ring and $I$ and $J$ are two ideals of $R$ .

Under what conditions it's true that $R/(I+J) \cong (R/I)/J \cong (R/J)/I$? (Please check my partial argument at the below). I think it's not true in general because $\mathbb{Z}/(3\mathbb{Z}+5\mathbb{Z})\not\cong (\mathbb{Z}/3\mathbb{Z})/5\mathbb{Z}$

Is it still true for non-commutative rings assuming that $I$ and $J$ are ideals of $R$ from right and left?

I think I should define $\varphi: R/J \to R/(I+J)$ by something like $\varphi(r+J)=r + (I+J)$.

I need to show that $\varphi$ is well-defined. So, if $r_1+J=r_2+J$ then $r_2-r_1 \in J$ and since $J \subset I+J$ we have $r_2-r_1 \in I+J$ which gives $r_2 + (I+J) = r_1 + (I+J)$, i.e. $\varphi(r_2 + J)=\varphi(r_1+J)$.

$\varphi$ is a homomorphism because:

  1. $\varphi((r_1+J) + (r_2+J))=\varphi((r_1+r_2)+J)=(r_1+r_2)+(I+J)= (r_1 + (I+J)) + (r_2+(I+J))=\varphi(r_1+J)+\varphi(r_2+J)$
  2. $\varphi((r_1+J) . (r_2+J))=\varphi(r_1.r_2 + J)=r_1.r_2 + (I+J) = (r_1+(I+J)) + (r_2+(I+J))=\varphi(r_1+J).\varphi(r_2+J)$

$\varphi$ is surjective. I think that's trivial from definition.

The kernel of $\varphi$ is as follows:

$\ker{\varphi} = {r+J: \varphi(r+J)=I+J}$.

There is no doubt that $I \subseteq \ker(\varphi)$. But the other direction looks confusing and hard. Do I have to assume $I\cap J=0$? or do we need other conditions or not?


If $I\cap J=0$ then since $I+J/J \cong I/(I\cap J) \cong I$ then $(R/J)/I \cong (R/J)/((I+J)/J) \cong R/(I+J)$. Is it the only way these two could get isomorphic?

  • 4
    $\begingroup$ How do you view $J$ as an ideal of $R/I$? The smallest ideal of $R/I$ that contains the cosets of all the elements of $J$ is $(I+J)/I$. With that correction your claim reads $$R/(I+J)\cong (R/I)/(I+J/I),$$ which is true. $\endgroup$ Sep 18, 2013 at 6:24
  • 2
    $\begingroup$ And in the counterexample $3\Bbb{Z}+5\Bbb{Z}=\Bbb{Z}$. Also $5$ is a unit in $\Bbb{Z}/3\Bbb{Z}$ so... $\endgroup$ Sep 18, 2013 at 6:26
  • $\begingroup$ @JyrkiLahtonen: Thanks. But I guess I still have understood something incorrectly. It's true that $<S \cup T>=<S>+<T>$, and it makes complete sense when $T$ and $S$ are finite. Now please check here: math.stackexchange.com/questions/355902/…. Can't we say $(I,a_1,\cdots,a_n)=<I>+<a_1>+\cdots+<a_n>$? Then isn't that isomorphism wrong if $<a_i> \cap <a_j> \neq 0$? I'm confused. I'd appreciate your help really. $\endgroup$
    – user66733
    Sep 18, 2013 at 9:10
  • $\begingroup$ In Alex Youcis' answer $I\subset R$ and $a_i\in R[X]$. He didn't state that explicitly, but it can be infered. Otherwise $R/I$ doesn't make sense. But I don't see why the isomorphism would be wrong, if $\langle a_i\rangle\cap\langle a_j\rangle\neq0$? Don't we always have $a_ia_j$ that intersection? In an integral domain two principal ideals always intersect non-trivially. For example in $\Bbb{Z}$ we have $3\Bbb{Z}\cap5\Bbb{Z}=15\Bbb{Z}$? So something is bothering you, but I can't figure it out yet, sorry. $\endgroup$ Sep 18, 2013 at 16:39
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    $\begingroup$ @JyrkiLahtonen: Ah.. my bad. I didn't read Hurkyl's answer carefully. I thought we needed to assume that $I \cap J = 0$ but that's not really required. Now the paradox is solved. Thank you for your help. $\endgroup$
    – user66733
    Sep 18, 2013 at 16:58

1 Answer 1


The third isomorphism is the way to go.

It may help to define things a bit more pedantically. If we're being picky and literal, $(R/J)/I$ doesn't actually make sense the way the notation is usually defined: $I$ is an ideal of $R$, not of $(R/J)$. What we really mean is $(R/J)/\pi(I)$, where $\pi : R \to R/J$ is the canonical projection.

It shouldn't be too hard to show that $\pi(I) \cong (I+J) / J$: this is essentially what you wanted to do with the second isomorphism theorem, except we no longer need to make an assumption on $I \cap J$.

  • $\begingroup$ By $\pi(I)$ you mean the extension of $I$ in $R/J$. Am I right? $\endgroup$
    – user66733
    Sep 18, 2013 at 18:01
  • $\begingroup$ @some1: Yes. Technically, I meant the direct image of the subset $I \subseteq R$ by the map $\pi$, but that set is already an ideal. $\endgroup$
    – user14972
    Sep 18, 2013 at 19:15
  • $\begingroup$ Would you please explain to me what you mean by direct image? and why $\pi(I)$ is already an ideal of $R/J$ without the need to extend it to an ideal? $\endgroup$
    – user66733
    Sep 18, 2013 at 19:18
  • $\begingroup$ If $f : X \to Y$ and $S \subseteq X$, then $f(S) := \{ f(s) \mid s \in S \}$. Sometimes this is written $f_*(S)$ instead, if there is a need to carefully distinguish between $f$ and $f_*$. $\endgroup$
    – user14972
    Sep 18, 2013 at 19:23
  • $\begingroup$ But that isn't necessarily an ideal of the co-domain set. Right? If $j: \mathbb{Z} \to \mathbb{R}$ then $j(2\mathbb{Z})$ is not an ideal of $\mathbb{R}$. I'm asking why $\pi(I)$ is an ideal of $R/J$ without extending it. If I've understood you correctly so far. $\endgroup$
    – user66733
    Sep 18, 2013 at 19:26

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