Let $R$ be a commutative Noetherian ring (with unity), and let $I$ be an ideal of $R$ such that $R/I \cong R$. Then is $I=(0)$? 
Let $R$ be a commutative Noetherian ring (with unity), and let $I$ be an ideal of $R$ such that $R/I \cong R$. Then is it true that $I=(0)$ ?

I know that a surjective ring endomorphism of a Noetherian ring is also injective, and since there is a natural surjection from $R$ onto $R/I$ we get a surjection from $R$ onto $R$, but the problem is I can not determine the map explicitly and I am not sure about the statement. Please help. Thanks in advance.

Assume $A\approx A/a$. Of those $A$ ideals $a'$ satisfying $A\approx A/a'$ choose $b$ maximal with respect to set inclusion. Consequently we may choose a non-trivial proper ideal $C$ of the the quotient ring $A/b$ such that $A\approx (A/b)/C$ but then $$A\approx (A/b)/C\approx A/(b+c)$$ where $c:=\{x\in A:x+b\in C\}$ is the pullback $A$ ideal of $C$, which then violates the maximality of $b$ because $C$ is non-trivial and thus $b+c$ would properly contain $b$.   
