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I think this is an obvious question, yet I'm stuck: Does $R/I \cong R$ implies $I=0$ where $R$ is a commutative ring with unity?

Say there exists $I\neq 0$ s.t. $R/I\cong R$. Then since $R$ is free, we have $R\cong R\oplus I$. I feel like this is a contradiction somehow. My intuition is that $R\oplus I$ cannot be generated by 1 element, but I cannot prove this rigorously.

Just a more general question: Does $R^2/I \cong R^2$ implies $I=0$?

T C
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    This has been answered here many times, consider $\mathbb{Z}^{\infty}$ – Tsemo Aristide Sep 07 '20 at 21:16
  • @TsemoAristide from context the OP is using $\cong$ to denote isomorphism of modules, not rings. – tkf Sep 07 '20 at 21:35
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    The vote to close refers to another question that is not a duplicate of this one (and has no accepted answer, but no-one seems to care about that these days). – Rob Arthan Sep 07 '20 at 22:42

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For a commutative ring $R$, you cannot have an isomorphism of modules $$R\cong R\oplus I,$$ with $I\neq 0$.

Proof: Given $a\in R$ which freely generates the summand $R$ and $b\neq 0$ with $b\in I$, we have $ab\in R\cap I$ so $ab=0$ (here by $R$ we mean the summand).

However $a$ freely generates the summand $R$, so $ab\neq 0$, a contradiction.

Given a surjective map $R^n\to R^n$, we may write it as a matrix $A$. Then as the map is surjective, we can find a matrix $B$ with $AB=I_n$ (use surjectivity to pick columns of $B$ one at a time). Then we have $$1=\det(AB)=\det(A)\det(B),$$ so $\det(A)$ is a unit in $R$ with inverse $\det(B)$. Thus $\det(B){\rm adj}(A)$ is the inverse of $A$, and the original surjective map has trivial kernel.

Now if $M$ is a submodule of $R^n$ and $R^n/M\cong R^n$, then we have a surjective map $$R^n\to R^n/M\cong R^n,$$ with kernel $M$. Thus $M=0$.

tkf
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  • This is great. But I tried and didn't think that the same method can apply to $R^2/I\cong R^2$? I mean this method is a clever way to use the fact that $I$ is an ideal of $R$. However, what we know is that $R^2/I\cong R^2$ as $R$-module and not $R^2$-module. You have an idea how to do the more general problem? – T C Sep 07 '20 at 21:56
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    For general $R^n$ the second answer in the link provided by Viktor Glombik should work. Also https://math.stackexchange.com/questions/239364/surjective-endomorphisms-of-finitely-generated-modules-are-isomorphisms. I will get back to you if I think of a simple proof in the $n=2$ case, but I expect that it would just end up being one of the proofs in those links. As you saw the $n=1$ case had a two line proof, so was worth mentioning separately. – tkf Sep 07 '20 at 22:14
  • Your answer doesn't make sense. How do you prove the non-existence of an isomorphism without even mentioning the isomorphism? – Rob Arthan Sep 07 '20 at 22:32
  • @RobArthan The isomorphism was implied, and meant I could regard elements of the summands as elements of the ring. – tkf Sep 07 '20 at 22:46
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    @TC I made an edit to cover general $n$ - I am sure at least one of the answers in the other posts uses this method, but it saves you having to hunt it down. – tkf Sep 07 '20 at 23:04
  • thanks you @tkf – T C Sep 07 '20 at 23:23