# When is $I/I^2$ isomorphic to $R/I$?

Let $R$ be a commutative ring with 1, and let $I$ be an ideal of $R$. What are some necessary and sufficient conditions that $I/I^2\cong R/I$? (Or is it always true?)

For instance, when $R=\mathbb{Z}$, and $I=p\mathbb{Z}$, we have that $$p\mathbb{Z}/p^2\mathbb{Z}\cong \mathbb{Z}/p\mathbb{Z}$$

Thanks for any help.

(I am looking at $R$-module isomorphism I suppose, since $p\mathbb{Z}/p^2\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/p\mathbb{Z}$ as rings since one has multiplicative identity and one doesn't.)

• Do you mean these isomorphisms as $R$-module isomorphisms? – Milo Brandt Nov 14 '17 at 4:07
• It's certainly not always true: take $R = k[x,y]$, $I = (x,y)$. Then $R/I$ is a cyclic $R$-module (generated by $1$), but $I/I^2$ cannot be generated by $1$ element. (Assuming you mean as $R$-modules.) I think assuming $I$ is locally free of rank $1$ is sufficient, at least. – André 3000 Nov 14 '17 at 4:20
• Aside: the $R$-module structure on these also induces an $R/I$-module structure. It's somewhat common to consider these as $R/I$-modules, especially when $I$ is a maximal ideal. – user14972 Nov 14 '17 at 5:13
• Counterexample: $R=F_2$ and $I=\{0\}$. – rschwieb Nov 14 '17 at 20:50

From your example I thin you mean “isomorphic as $R$ modules.”
Well, it’s obviously not always true. If $I\neq R$ and $I^2=I$, the second quotient will be zero and the first will not.
Another way it can fail is if $I/I^2$ is not cyclic. For example, take any noncyclic abelian group $G$, and give it zero multiplication, and take the trivial extension by $\mathbb Z$. Then $I/I^2\cong I$ is not cyclic.
The example you gave can be generalized slightly: it works for any element $p$ which isn’t a zero divisor of $R$ because the obvious map $pr\mapsto r$ is well defined.
If $R/I^2$ is an artinian $R$ module, it also seems like it’s necessary for its composition factors to occur in pairs.
A necessary and sufficient condition would be for there to exist an element $p\in I$ such that $pR/I^2=I/I^2$ and $\{x\in R\mid px \in I^2\}=I$. That’s not very interesting to look at but there you have it.