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 $I=0$?
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 $M=0$.