Let $A$ be Noetherian ring that is an integral domain and let $\frak a$ be a proper ideal.
I would like to know if it can happen that $\cap\_{1}^{\infty}{\frak a}^n\ne 0$ and at the same time the sequence of ideals ${\frak a}^n$ does not stabilise? What would be a "natural example"?


(this question is motivated by trying to understand Krull intersection theorem).


PS. I was thinking of the following version of Krull intersection theorem:


**Theorem.** Let a $\frak a$ be be an ideal in a noetherian
ring $A$. If $\frak a$ is contained in all maximal ideals of $A$, then $\cap\_{1}^{\infty}{\frak a}^n=0$.



Infinite power of in ideal in a Noetherian ring
If $P$ is a prime ideal containing $a$ then $\cap a^n$ is contained in $\cap P^n$, whose image in $A\_P$ vanishes, so $\cap a^n$ has vanishing image in $A\_P$ for every point $P$ of Spec($A/a$). Such $P$ exist as long as $a$ isn't the unit ideal, so if $A$ is an integral domain then $\cap a^n = 0$ in $A$ since $A \rightarrow A\_P$ is injective in such cases.


In contrast, for $A = k[x,y]/(xy) = k[x] \times\_k k[y]$ for a field $k$, and 
$$a = (x-1)A = (x-1)k[x] \times\_k k[y],$$ we have $\cap a^n = yk[y] \ne 0$ and $\{a^n\}$ doesn't stabilize. Thus, the "domain" condition cannot be relaxed to mere connectedness of the spectrum (even assuming reducedness).


[${\bf{Remark}}$: An earlier version of this answer had the following bogus "proof" that $\cap a^n = 0$ assuming that Spec($A$) is merely connected, and some of the comments refer to this bogus argument. So here it the incorrect proof of that incorrect generalization, with the mistake in the reasoning identified. 


"Proof": Arguing as above, since $\cap a^n$ has vanishing image in the stalk $A\_P$ at each point of Spec($A/a$), it vanishes over an open neighborhood $U$ of this closed set. On the other hand, the ideal *sheaf* $J = \cap (\widetilde{a})^n$ is the unit ideal on the open set $V$ of Spec($A$) complementary to the closed set Spec($A/a$). Since $\cap a^n$ is the ideal of global sections of $J$, if $J$ is *quasi-coherent* (equivalently, coherent) then $U$ and $V$ are disjoint open sets that cover the entire space, which would imply by connectedness that either $V$ is empty (in which case $\cap a^n = 0$ in $A$) or $U$ is empty (in which case $a$ is the unit ideal). "QED"


The error is that often the ideal sheaf $\cap (\widetilde{a})^n$ is *not quasi-coherent* (equivalently, isn't coherent), which is to say that it doesn't correspond to its ideal of global sections $\cap a^n$. This is an example of the fact that quasi-coherence can be destroyed by inverse limits, and more concretely inverse limits don't naturally commutes with tensor product or even just localization in general. 


The failure of coherence for this ideal sheaf is seen rather concretely in the case $A = k[x,y]/(xy)$ with $a = (x-1)$ since the sheaf $\cap \widetilde{a}^n$ is the unit ideal sheaf away from the point $(1,0)$ whereas its ideal of global sections $\cap a^n = y k[y]$ is the ideal of the $x$-axis and so does not localize to the unit ideal at points $(c,0)$ with $c \ne 1$ (e.g., $c = 0$). This illustrates that the coherent ideal sheaf associated to $\cap a^n$ can be rather smaller than $\cap \widetilde{a}^n$.]


