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 $I=(0)$?

Question

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.

Answer 1

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 $I = \{0\}$

Answer 2

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).