counterexample related to Chevalley's Theorem 
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?
 A: 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 $\mathfrak{m}^k$.     
A: 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 $n$.
