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.
 A: Assume that $R/I$ and $R$ are isomorphic. Let us denote the isomophism by $f:R/I \to R$. 
Let $\pi:R \to R/I$ denote the usual map $x \mapsto x + I$. 
This is a of course a surjective ring homomorphism. 
The composition $f \circ \pi : R \to R$ is thus a surjective ring endomorphism (the composition of surjections is a surjection). 
By the result quoted in the question $f \circ \pi$ is an isomorphism, in particular it is injective. It follows that $\pi$ is injective, otherwise the composition could not be injective. 
The kernel of $\pi$ is thus $\{0\}$; it is also  is $I$. Thus $I = \{0\}$
A: Assume $f:R/I\to R$ is an isomorphism and $I \ne (0)$. Let $\overline{J} = f^{-1}I\subset R/I$ and $J\subset R$ be the preimage of $\overline J$ in $R$. Now $I \ne (0)$ implies $J$ strictly contains $I$; but $R/J \cong (R/I)/\overline J$ which is isomorphic to $R/I$ via $f$, and so isomorphic to $R$ by hypothesis. Now you can repeat for $J$; you will find a never-ending sequence of ever-larger ideals, contradicting the Noetherian property.
Note that the problem statement is somewhat ambiguous; you could interpret the isomorphism to be "isomorphic as $R$-modules" (in which case the problem would be trivial).
A: 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!
A: 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 $b$.   
