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Chevalley's Theorem:

$(A,m)$ : a complete local ring.

${a_n}$ : a descending sequence of ideals with $\bigcap a_n=(0)$

Then for each $k$, there is an $n(k)$ s.t. $n\ge n(k)$, $a_n\subset m^k$.

Can you give a counterexample if $A$ is not complete?

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The title is not appropriate, this would be no counterexample to Chevalley's theorem. – YCor Jun 20 '13 at 11:36
Following the above comment I changed the title. @Yves Cornulier: it is IMO frequently better to just make such a change (and document the reasoon via a comment) as opposed to just pointing out its necessity. (OP can still rollback or chose something still different if they want). – user9072 Jun 20 '13 at 13:42

I saw an example posted, but then it disappeared. Here is a counterexample (roughly the same as I remember seeing posted). Let $A$ be the local ring of $k[x,y]$ at the maximal ideal $\mathfrak{m} = \langle x,y \rangle$. Embed $A$ in its completion, $\widehat{A} = k[[x,y]]$. Let $p(x)$ be an element in $xk[[x]]$ that is not in algebraic over the local ring $k[x]_{\langle x \rangle}$, e.g., $p(x) = xe^x$. Form the ideal $I = \langle y - p(x) \rangle$ in $\widehat{A}$. For every integer $n>0$, let $\mathfrak{a}_n$ be the ideal in $A$ that is the intersection of $A$ with $I + \mathfrak{m}^n$. By Krull's Intersection Theorem, the intersection over all $n$ of $I+\mathfrak{m}^n$ is just $I$. Since $I\cap A$ equals $\{0\}$, it follows that the intersection over all $n$ of $\mathfrak{a}_n$ equals $\{0\}$. Yet for every $n$ and for every $k\geq 2$, $\mathfrak{a}_n$ is not contained in $\mathfrak{m}^k$.

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Let $R = k[x_0,\dots]$ a polynomial ring over a field with infinitely many variables, and put $\mathfrak{m} = (x_0,\dots)$ the ideal generated by degree one homogeneous polynomials(that is, the variables). Put $A = R_\mathfrak{m}$ and by abuse of notation just write $\mathfrak{m}$ for the maximal ideal of $A$.

Define $a_n = (x_0^{\max \lbrace n,0 \rbrace }, x_1^{\max \lbrace n-1,0 \rbrace }, x_2^{\max \lbrace n-2,0 \rbrace }, \dots, x_i^{\max \lbrace n-i,0 \rbrace }, \dots)$. That is, $a_1 = \mathfrak{m}$, $a_2 = (x_1^2,x_2,x_3,\dots ) $ and so on. Clearly $a_{n+1} \subsetneq a_n$, and $\cap_n a_n = (0)$. But also by construction for any $k>1$, no containment $a_n \subset \mathfrak{m}^k$ holds for any $n$.

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Your first example is fine (although non-Noetherian). Your easier example is not okay: the element $x_2$ is in each of those ideals. – Jason Starr Jun 20 '13 at 13:42
Yes, thank you, I'll remove it. – John Stalfos Jun 20 '13 at 15:04

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