TITLE: Is $\bigcap_{n=1}^\infty I^n$ idempotent, for any ideal $I$?
QUESTION [6 upvotes]: Let $I$ be an ideal in a ring $R$. By $I^n$, let us understand the ideal generated by all $n$-fold products $x_1x_2\cdots x_n$ where $x_1, \ldots, x_n \in I$. Obviously $I \supseteq I^2 \supseteq I^3 \supseteq \ldots$. Let us also define $I^\infty =\bigcap_{n=1}^\infty I^n$. 

Question: Does the ideal $I^\infty$ always satisfy $(I^\infty)^2 = I^\infty$?

If we naively consider elements of $I^\infty$ to be something like "infinitely long products", then it is tempting to think this should be true by some kind of infinite swindle type argument. However, if we are being honest, the elements of $I^\infty$ are the elements which can be expressed as (sums of) arbitrarily long products of elements of $I$. But there does not appear to be any compelling reason why each such element should be be expressible as a sum of products of such elements. 
Remark: If you like nonunital rings, then it is not necessary to introduce an ideal $I$ to pose this question. We can just work with powers of the ring $R$.

REPLY [2 votes]: For a more "natural" example, if you don't mind noncommutative rings, pick a field $k$ and consider the non-commutative ring $k[X^{\omega_1}]$ of polynomials over $k$ with exponents in $\omega_1$ (so that for any ordinals $\alpha$ and $\beta$, you have $X^\alpha\cdot X^\beta=X^{\alpha+\beta}$).
This is a local ring with the maximal ideal $\mathfrak m=(X)$. Note that $X^\omega=(X^1)^n\cdot X^\omega=X^{n+\omega}=X^\omega$ for all $n$, so $\mathfrak m^\omega:=\bigcap_{n\in \omega} \mathfrak m^n=(X^\omega)$, which is clearly not an idempotent (in fact, even $\mathfrak m^\omega \cdot \mathfrak m\neq \mathfrak m^\omega$).
In this manner, you can find ideals which take arbitrarily long to stabilise.