For a finitely generated ideal $I$, $I^2=I$ implies $A/I$ is flat over $A$ or $I$ is principal with an idempotent

Question

This is exercise 1.2.4 from Liu's Algebraic Geometry and Algebraic Curves

Let $I$ be a finitely generated ideal of a (commutative with $1$) ring $A$. The following are equivalent:

(a) $A/I$ is flat over $A$

(b) $I=I^2$

(c) There exists an $e\in A$ with $e^2=e$ and $I=eA$

After writing down the question and my concerns, I inevitably found the answer which I'm posting below. I still would like to know if there is a way to prove (b) implies (a), and if Liu's Nakayama's lemma (as stated below), is enough to prove (b) implies (c).

Let $(A,\mathfrak m)$ local, $M$ a finitely generated module and $M=\mathfrak m M$. Then $M=0$.

Answer 1

Since I spent at least half an hour writing this question down only to find the solution myself, I thought at least I'd post it to not have it be time completely wasted.

(a) implies (b) is easy, by tensoring the exact sequence $I\hookrightarrow A$ with $A/I$. (c) implies (b) is trivial. (c) implies (a) is equivalent to $\phi:N\hookrightarrow M$ injective implies $N/eN\hookrightarrow M/eM$ injective. If $\phi(n)=em$, this follows from $\phi(en)=e^2m=em=\phi(n)$, implying $n=en\in eN$ by injectivity.

Painfully, I found out that Liu's Nakayama Lemma is not enough (or at least not obvious) to hint at (b) implies (c). Matsumura's Nakayama says:

Let $A$ a ring, $I$ an ideal and $M$ a finitely generated module with $IM=M$. Then there exists an $r\in A$ with $r\equiv 1 \pmod I$ and $rM=0$.

Now suppose $I^2=I$. The latter implies that $\exists r$ with $rI=0$ and $e:=1-r\in I$. The canonical element to choose would be $e$, since $r\not\in I$ unless $I=A$, and from $rI=0$ we get $r(r-1)=0$, ie $r^2=r$. Then $e^2 = 1-2r+r^2=1-r=e$, as required. But now for any $a\in A$, $a= ae + a(1-e)$ and $(e)\cap (1-e)=0$, so $A=(e)\oplus (1-e)$ and since $e\in I$ and $1-e\not\in I$, then $I=eA$.

Answer 2

Here is a proof that (b) implies (c) using just the local Nakayama lemma. Suppose $I^2=I$ and let $P\subset A$ be any prime ideal containing $I$. Localizing at $P$, we find that $I_P=I_P^2\subseteq P_PI_P$ and so $I_P=0$ by Nakayama. This means (again using the fact that $I$ is finitely generated) there exists some $s\in A\setminus P$ such that $sI=0$.

Now let $J=\{s\in A:sI=0\}$. By the discussion above, $J$ is not contained in any prime ideal containing $I$. That is, $I+J$ is not contained in any prime ideal, so $I+J=A$. Now pick $e\in I$ and $f\in J$ such that $e+f=1$. For any $i\in I$, we then have $$i=(e+f)i=ei+fi=ei$$ since $fI=0$. This implies that $e$ generates $I$, and also that $e^2=e$ by taking $i=e$, as desired.