Is $\bigcap_{n=1}^\infty I^n$ idempotent, for any ideal $I$?

Question

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 $R$.

Answer 1

No, this is not true. For instance, let $k$ be your favorite field, let $R$ be the subring $k[x,y,x^{-1}y,x^{-2}y,x^{-3}y,\dots]\subset k(x,y)$ and let $I$ be the ideal generated by $x$. Then $I^\infty$ is the ideal generated by $x^{-n}y$ for all $n\in\mathbb{N}$. In particular, $y\in I^\infty$, but $y\not\in (I^\infty)^2$ since every element of $(I^\infty)^2$ is divisible by $x^{-n}y^2$ for some $n$.

In fact, it is not even necessarily true that $I\cdot I^\infty=I^\infty$. For instance, let $R=k[y,x_1,x_1^{-1}y,x_2,x_2^{-2}y,x_3,x_3^{-3}y,\dots]$ and let $I$ be the ideal generated by all the $x_n$'s. Then $y\in I^\infty$ since it is divisible by $x_n^n$ for each $n$, but $y\not\in I\cdot I^\infty$, essentially because once you divide $y$ by a single $x_n$ you lose divisibility by all the others (this takes some work to turn into a rigorous proof, though).

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You can even get a Noetherian counterexample. Let $R=k[x,y]/(xy-y)$ and let $I=(x)$. Then $I^\infty=(y)$, which is not equal to its square. In a Noetherian (commutative) ring though we must at least have $I\cdot I^\infty=I^\infty$, by the Artin-Rees lemma.

Answer 2

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.