Infinite Intersection of ideals generated by idempotents Let $R$ be a commutative ring with unity. Let $\{I_{\alpha},\alpha\in \Delta\}$ be a set of ideals such that for each $\alpha\in\Delta$, $I_\alpha$ is an ideal generated by idempotents of $R$.
Does $J=\bigcap\limits_{\alpha\in \Delta} I_\alpha$ is an ideal finitely generated by idempotents when $\Delta$ is countable or finite set? 
If no, does $J$ is an ideal generated by idempotents when $\Delta$ is a finite set?
 A: In the countable case, $J$ need not be generated by idempotents.  For instance, let $X=[-1,0]\cup\{1/n:n\in\mathbb{Z}_+\}$ and let $R$ be the ring of continuous functions $X\to\mathbb{R}$.  For each $n$, let $I_n$ be the ideal of functions that vanish at $1/n$.  Each $I_n$ is generated by an idempotent, namely the characteristic function of $X\setminus\{1/n\}$.  But $J=\bigcap_n I_n$ is not generated by idempotents.  Indeed, if $f\in J$ is idempotent, then $f$ vanishes at $1/n$ for each $n$, and hence also at $0$ by continuity.  Since $f$ can only take the values $0$ and $1$, it must thus vanish on all of $[-1,0]$ as well.  So the only idempotent in $J$ is $0$, but $J$ certainly contains some nonzero functions (for instance, the function that is the identity on $[-1,0]$ and $0$ at each $1/n$).
On the other hand, a finite intersection of ideals generated by idempotents is generated by idempotents.  It suffices to consider binary intersections; suppose $I_0$ and $I_1$ are both generated by idempotents, and let $f\in I_0\cap I_1$.  Since $I_0$ is generated by idempotents, there is some idempotent $e_0\in I_0$ such that $e_0f=f$ (namely, take finitely many idempotents in $I_0$ that generate $f$, and let $e_0$ be their join).  Similarly there is an idempotent $e_1\in I_1$ such that $e_1f=f$.  Now let $e=e_0e_1$; then $e\in I_0\cap I_1$ is idempotent and $ef=f$.  Thus $f$ is generated by $e$.  Since $f\in I_0\cap I_1$ was arbitrary, $I_0\cap I_1$ is generated by idempotents.
