Intersection of powers of maximal ideals Let $A=\mathbb K[X_1,\ldots,X_n]$ be a polynomial ring over some field $\mathbb K$. Let $\mathfrak p\subseteq A$ be a prime ideal. Let $Z(\mathfrak p)=\{ \mathfrak m\subset A\text{ maximal}\mid \mathfrak p\subseteq\mathfrak m\}$ be the set of maximal ideals of $A$ that lie over $\mathfrak p$. My intuition says that 
$$ \bigcap_{\mathfrak m\in Z(\mathfrak p)} \mathfrak m^s=\mathfrak p^s$$
because instead of counting a subvariety $s$ times, I can count each of its points $s$ times. Is this true? If no, what if $\mathbb K$ is algebraically closed? If yes, does it hold for a certain  larger class of rings $A$?
 A: For $s=1$, as Martin Brandenburg correctly states, your condition is exactly that of being a Jacobson ring, and hence is true for any factor ring of a polynomial ring in finitely many variables over a field or over $\mathbb{Z}$.
However, for $s>1$, your condition should generally fail even in polynomial rings over an algebraically closed field.  Indeed, the Zariski-Nagata theorem says that if $k$ is an algebraically closed field, $R=k[X_1, \dotsc, X_n]$, and $\mathfrak{p} \in \operatorname{Spec} R$, then the intersection you give is the $s$th symbolic power of $\mathfrak{p}^{(s)}$ of $\mathfrak{p}$.  By definition $\mathfrak{p}^{(s)} = \mathfrak{p}^s R_{\mathfrak{p}} \cap R$, which in general can be larger than $\mathfrak{p}^s$.  In an article of Huneke in Math. Ann. from 1986, for example, he gives a way to construct 3-generated prime ideals $\mathfrak p$ of height two in $k[[X,Y,Z]]$ (which can easily be altered to give such examples in $k[X,Y,Z]$) such that $\mathfrak p^{(2)}$ has arbitrarily many generators.  (That is, pick a number; he can then give you a 3-generated prime ideal whose second symbolic power has at least that many generators.)  But it's obvious that any such $\mathfrak p^2$ can be generated by at most 9 elements (namely, the pairwise products of the original generators of $\mathfrak p$).
