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So as seen on this question Converse of the Chinese Remainder Theorem, we know that if $(n,m) \neq 1$, then $\mathbb{Z} /mn \mathbb{Z} \ncong \mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, because the right hand side does not have an element of order $nm$.

But take a more general setting. Let $R$ be a commutative ring, and let $A,B$ be ideals in $R$. $A$ and $B$ are said to be comaximal if $A+B=R$. If $A, B$ are comaximal, then we have: $R/AB \cong R/A \times R/B$. In this setup, is the converse true?

If we have $A, B$ ideals in $R$ such that $R/AB \cong R/A \times R/B$, do we always have that $A$ and $B$ are comaximal?

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MadMonty
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  • http://math.stackexchange.com/questions/1243503/if-r-i-times-r-j-is-isomorphic-to-r-i-cap-j-as-r-modules-then-i-j/1243574#1243574 – user26857 May 31 '15 at 00:20

2 Answers2

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I'll assume the isomorphism is as $R$-modules (or the result is false, see Can $R \times R$ be isomorphic to $R$ as rings?).

Suppose $R/A\oplus R/B\cong R/AB$, in particular is cyclic, and that $(x+A,y+B)$ is a generator. Then, for some $r\in R$, $(1+A,0+B)=r(x+A,y+B)=(rx+A,ry+B)$, so $$ 1=rx+a,\qquad ry\in B $$ for some $a\in A$. Then $y=y1=ay+rxy\in A+B$. If $z\in R$, then $$ (0+A,z+B)=s(x+A,y+B) $$ for some $s\in R$, which implies $z+B=sy+B\in(A+B)/B$, so $A+B=R$.

The same proof of course applies if instead of $AB$ we have any other ideal, in particular $A\cap B$.

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egreg
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    I really don't understand your motivation to post this answer: your remark about rings can be found in my answer, and the proof for modules is more or less similar to the one given here. – user26857 Apr 20 '15 at 23:25
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    @user26857 This proof is completely elementary and doesn't use tensor products or bilinear maps. I like my answer are complete; if you want you can remove the comment about the isomorphism as rings. – egreg Apr 20 '15 at 23:27
  • Bilinear maps are used there as a concept to show the OP the connection with tensor products and exterior powers, but it is not really necessary. As I said, the proofs are quite similar. – user26857 Apr 20 '15 at 23:29
  • @user26857 I see no point in even mentioning tensor products or bilinear maps, when there is an elementary proof. This might come later for showing the result from a higher point of view. I feel that using the existence of maximal ideals for proving this elementary fact is overkill. – egreg Apr 20 '15 at 23:32
  • @egreg $R/A\oplus R/B\cong R/AB$ cyclic? Other than that I like your answer – user10024395 Apr 21 '15 at 01:03
  • @egreg Why is it sufficient to suppose it's a cyclic module? – MadMonty Apr 21 '15 at 06:55
  • @Aha $R/I$ is generated by $1+I$, for every ideal $I$. Any module isomorphic to a cyclic module is cyclic. – egreg Apr 21 '15 at 10:26
  • @user1440894 That's the only used fact. – egreg Apr 21 '15 at 10:26
  • @egreg what happens if $z \in B$, then i can just have $sy \in B$.THen how do i ensure that z can be written as A+B? I don't quite understand the last part. Why $z+b = sy +B \in (A+B)/B, so A+B =R$ Can you explain more about the reasoning behind? Thanks – user10024395 Apr 21 '15 at 13:01
  • @Aha Correspondence theorem: if $I$ and $J$ are ideals containing the ideal $K$, then $I=J$ if and only if $I/K=J/K$. Or, more simply, if $z\in B$, then of course $z\in A+B$. – egreg Apr 21 '15 at 13:09
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If you are considering the quotients as $R$-modules, it's not hard to prove that if $I,J$ are not comaximal, then $R/IJ\not\simeq R/I\times R/J$: take a maximal ideal $M$ containing $I,J$ and tensor the both sides by $R/M$ (over $R$) in order to get a contradiction. (In fact, $R/I\times R/J$ is a cyclic $R$-module iff $I+J=R$.)

If the isomorphism is of rings, then the claim holds trivially for local rings. For the general case consider a ring $R$ such that $R\simeq R\times R$ (for such an example look here) and $I=J=(0)$.

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