When an ideal is locally comaximal with idempotents(restated)

Question

I saw the following question at mathstackexchang < https://math.stackexchange.com/questions/2282194/when-an-ideal-is-locally-comaximal-with-idempotents>. It seems to be a nice question and I need it in my research. I want to know is it true? It is clearly true for indecomposable rings.

Let $R$ be a commutative ring with identity, and let for an ideal $A$ of $R$, $\mathrm{Id}(A)$ be the ideal of $R$ generated by all idempotent elements of $A$. Now let $I$ be an ideal of $R$ such that for each prime ideal (or primary ideal) $P$ of $R$ we have $I\subseteq \mathrm{Id}(P)$ or there exists $a^2=a\in P$ with $a-1\in I$. Show that $I=0$.

Answer 1

No,this is false. Take $R=\mathbb{C}\times\mathbb{C}$. The primes ideals of $R$ are $P_1=\mathbb{C}\times (0)$ and $P_2=(0)\times \mathbb{C}$. The idempotents in $P_1$ are $(0,0)$ and $(1,0)$, and those in $P_2$ are $(0,0)$ and $(0,1)$. Hence $I(P_1)=P_1$ and $I(P_2)=P_2.$

Now take $I=P_1=\mathbb{C}\times (0)$. Then $I\neq (0)$. However, $I=P_1\subset I(P_1)=P_1$, and $a=(0,1)$ is an idempotent of $P_2$ such that $a-1_R=(-1,0)\in \mathbb{C}\times (0)=I$.