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 $a$ is a proper ideal. Suppose they were isomorphic. Then $\varphi: A \to A/a$ is some arbitrary isomorphism, and correspondingly $\varphi(a) := I_{1} \subset A/a$ is an ideal (proper inclusion as $a \subset A$ is a proper inclusion). By the correspondence principle, $I_{1} \subset A/a$ pulls back to an ideal $a \subset I'_{1} \subset A$ which are all proper inclusions (as $\varphi(a)$ is nonzero in $A/a$), where $I'_{1} = \pi^{-1}(I_{1})$ for $\pi: A \to A/a$ the canonical projection map. 
The key step is that $\varphi$ induces an isomorphism $\overline{\varphi}: A/a \to A/I_{1}$. But then $\overline{\varphi}(I_{1}):= I_{2}$ pulls back to an ideal $I'_{2} \subset A$ (proper inclusion using the same rationale as above) via $\pi_{1}: A \to A/I_{1}$ the canonical projection, such that $a \subset I'_{1} \subset I'_{2}$ in $A$ are all proper inclusions. Now iterate this process to yield an ascending chain which does not stabilize. Contradiction!
