Counter-examples to Krull's intersection theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:15:31Z http://mathoverflow.net/feeds/question/71699 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71699/counter-examples-to-krulls-intersection-theorem Counter-examples to Krull's intersection theorem Gengis Khan 2011-07-31T02:24:47Z 2011-08-04T16:01:42Z <p>The more general form of Krull intersection theorem says:</p> <blockquote> <p>Let $R$ be local and Noetherian and $I \subset R$ a proper ideal. If $M$ is finitely generated over $R$, and $N=\cap_1^{\infty} I^iM$, then $IN=N$. </p> </blockquote> <p>What is the simplest counter-examples when one (and only one) condition among: $R$ local, $R$ Noetherian or $M$ finitely generated is dropped? So this is three questions I guess. </p> <p>Sorry if this is too easy for this site. It has been a while, you know! </p> <p>LATER: Andrea's answer gave a counter-example to the stronger statement: there is an element $r \in I$ such that $N(1-r)=0$. I believe it is not a counter-example to the form stated above, see David Eisenbud's book on commutative algebra, the Example after Corollary 5.5 and Exercise 5.6.</p> <p>However, Dustin Cartwright pointed out that one can safely drop the "local" hypothesis. So there are only two questions left. </p> http://mathoverflow.net/questions/71699/counter-examples-to-krulls-intersection-theorem/71718#71718 Answer by Andrea for Counter-examples to Krull's intersection theorem Andrea 2011-07-31T09:10:59Z 2011-07-31T09:10:59Z <p>If $R$ is not Noetherian, you could read remark 2 at page 110 of <em>Atiyah, Macdonald - Introduction to Commutative Algebra</em>.</p> http://mathoverflow.net/questions/71699/counter-examples-to-krulls-intersection-theorem/71987#71987 Answer by a-fortiori for Counter-examples to Krull's intersection theorem a-fortiori 2011-08-03T11:22:49Z 2011-08-03T20:09:49Z <p>For $R$ noetherian and $M$ not finitely generated you can take the following example from Kaplansky, <em>Infinite Abelian Groups</em>: The abelian group $G$ with generators $x$ and $y_k$ for $k=1,2,\dots$ and relations $px=0$, $x=py_1=p^2y_2=\dots=p^ky_k=\dots$ ($p$ some fixed prime) satisfies $G_\omega=\bigcap p^kG=\langle x\rangle$, but $pG_\omega=0\ne G_\omega$.</p> <p>Building upon this example, you also get an example for the case $R$ non-noetherian and $M$ finitely generated: with the same $G$, you can set $R=\mathbb Z\times G$ with $G$ as a square-zero ideal, $M=R$ and $I=pR$. Now, $\bigcap I^n=0\times G_\omega$ again satisfies $I\bigcap I^n\ne\bigcap I^n$.</p> http://mathoverflow.net/questions/71699/counter-examples-to-krulls-intersection-theorem/72093#72093 Answer by Dustin Cartwright for Counter-examples to Krull's intersection theorem Dustin Cartwright 2011-08-04T16:01:42Z 2011-08-04T16:01:42Z <p>Here's a variation on a-fortiori's answer Let $R=\mathbb Q[x, z, y_1, y_2, \ldots]/ \langle x-zy_1, x-z^2y_2, \ldots \rangle$. Then take $M=R$ and $I=\langle z \rangle$, and then $\cap_{i=1}^n I^n = \langle x \rangle$, but $z\langle x \rangle \neq \langle x \rangle$. Obviously, in this case, $R$ is not Noetherian.</p> <p>Now keep $M$ as above, but replace $R$ with the subring $\mathbb Q[z]$. The intersection theorem fails in the same way as above. In this case, $R$ is Noetherian, but $M$ is not a finitely generated $R$-module.</p>