Infinite power of in ideal in a Noetherian ring - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:49:03Z http://mathoverflow.net/feeds/question/109040 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109040/infinite-power-of-in-ideal-in-a-noetherian-ring Infinite power of in ideal in a Noetherian ring aglearner 2012-10-07T05:22:49Z 2012-11-05T13:59:18Z <p>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"?</p> <p>(this question is motivated by trying to understand Krull intersection theorem).</p> <p>PS. I was thinking of the following version of Krull intersection theorem:</p> <p><strong>Theorem.</strong> 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 $\cap_{1}^{\infty}{\frak a}^n=0$.</p> http://mathoverflow.net/questions/109040/infinite-power-of-in-ideal-in-a-noetherian-ring/109041#109041 Answer by xbnv for Infinite power of in ideal in a Noetherian ring xbnv 2012-10-07T05:37:06Z 2012-11-05T13:59:18Z <p>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.</p> <p>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 <code>$\{a^n\}$</code> doesn't stabilize. Thus, the "domain" condition cannot be relaxed to mere connectedness of the spectrum (even assuming reducedness).</p> <p>[${\bf{Remark}}$: 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. </p> <p>"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 <em>sheaf</em> $J = \cap (\widetilde{a})^n$ 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 <em>quasi-coherent</em> (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"</p> <p>The error is that often the ideal sheaf $\cap (\widetilde{a})^n$ is <em>not quasi-coherent</em> (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. </p> <p>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 $\cap \widetilde{a}^n$.]</p>