# 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 $A$?

• You mean that contain $\mathfrak{p}$? Apr 30, 2013 at 9:50
• Yes, those that contain $\mathfrak p$.
– user38451
Apr 30, 2013 at 9:53
• +1. Remark: For $s=1$ this is precisely the Jacobson property. Apr 30, 2013 at 10:08

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 $\mathfrak p$).

• Nice! Zariski-Nagata Theorem was the result I was looking for, since I was convinced that in some conditions the LHS is a symbolic power. The only thing I'd like to add to your answer is another reference to an example of prime ideal whose second power differs from its second symbolic power: it is the celebrate example of Macaulay.
– user26857
Apr 30, 2013 at 20:10
• Thanks for the simple example. It is the typical example of a non complete intersection curve. If $K$ is perfect, $\mathrm{Spec}(R)$ is a smooth algebraic variety over $K$ and if $\mathfrak p$ is locally defined by a regular sequence (i.e. $\mathrm{Spec} (R/\mathfrak p)$ is locally complete intersection), then one can show $\mathfrak p^s$ is actually equal to the expected intersection of powers of maximal ideals.
– user18119
May 1, 2013 at 4:09
• @QiL'8: Do you have a reference for that last fact? Cases in which equality actually holds are very interesting for me.
– user38451
May 1, 2013 at 14:32
• @QiL'8 If you are interested, the paper of Huneke has the title The primary components of and integral closures of ideals in 3- dimensional regular local rings and is open access via digizeitschriften.de/en
– user26857
May 1, 2013 at 16:39
• @Karl: this should be well-known. I just did some computations by myself. For example, in the case $s=2$, if $f$ belongs to the intersection of $m^2$, then its differential $df$ vanishes at all points $V(\mathfrak p)$. Using the case $s=1$ and the fact that cotangent sheaf of the ambient space is locally free, we see that $df$ has coefficients in $\mathfrak p$. "Integrating $df$" then implies that $f\in \mathfrak p^2$. Higher powers are more complicated especially in positive characteristics. Note that locally complete intersection subvarieties include smooth subvarieties...
– user18119
May 1, 2013 at 16:41