Let $I\subseteq{\mathbb C}[X\_1,\dotsc,X\_n]$ be an ideal, and


let $V\subseteq{\mathbb C}^n$ be the corresponding algebraic set


($V$ consists of those $x$ at which all $f\in I$ vanish).


Is it true that then there exists an integer $N$ such that:


if a function $f\in{\mathbb C}[X\_1,\dotsc,X\_n]$
 and all its partial derivatives up to order $N$
 vanish on $V$,


then $f\in I$?


MOTIVATION. I believe this would imply an affirmative answer to my question about commutative algebras and representations of the category of finite sets:


[Commutative algebras and Gamma-modules](https://mathoverflow.net/questions/41440/commutative-algebras-and-gamma-modules)



Strong Nullstellensatz
The answer is yes. Let $I = q\_1 \cap q\_2 \cap \cdots \cap q\_k$ be the primary decomposition of $I$. Let $p\_i$ be the radical $\sqrt{q\_i}$ and let $N\_i$ be large enough that $q\_i \supseteq p\_i^{N\_i}$. I claim that we can take $N= \max(N\_i)$. We'll abbreviate $\mathbb{C}[x\_1, \ldots, x\_n]$ to $A$.


Let $f$ be a polynomial as in the statement of the question. In order to show that $f \in I$, it is enough to show that $f \in q\_i$ for every $i$. We focus on one $q$ to pay attention to, so we can drop the subscript $i$ and just talk about $p$, $q$ and $N$. Let $W$ be the variety of $p$.


Now, $A$ is regular, so $A\_p$ is regular. In other words, if $p$ has codimension $d$, there are transverse smooth hypersurfaces $y\_1$, $y\_2$, ..., $y\_d$ which generate the maximal ideal of $A\_p$. So there is some $u \not \in p$ such that $u^{-1} p = \langle y\_1, \ldots, y\_d \rangle$ in the localization $A[u^{-1}]$.


We claim that $f$, as an element of $A[u^{-1}]$ is in $u^{-1} p^N$. Proof: Write 
$$f=\sum\_{i\_1+\cdots+i\_d < N} b\_{i\_1 \ldots i\_d} y\_1^{i\_1} \cdots y\_d^{i\_d} + r$$
with $r \in u^{-1} p^N$ and with all the $b$'s either $0$ or not in $p$. We want to show the $b$'s are zero. If not, let $(i\_1, \ldots, i\_d)$ be minimal with the corresponding $b$ nonzero. Since $b$ is not in $p$, it is not zero on $W$. Choose $w$ in $W$ where neither $b$ nor $u$ vanishes. 


Since the $y\_i$'s are transverse hypersurfaces, we can take $\sum i\_j$ derivatives to obtain a polynomial with leading term $\mbox{nonzero stuff} \cdot \prod (i\_j)! \cdot b$ at $w$. Since we are in characteristic zero, this doesn't vanish at $w$. (In finite characteristic, the factorials might be zero.) But $w \in W \subseteq V$, so this derivative is supposed to vanish at $w$. This contradiction shows that $f \in u^{-1} p^N$.


Since $q \supseteq p^N$, we have $f \in u^{-1} q$ so $u^k f \in q$ for some $k$. But $q$ is primary and $u \not \in p = \sqrt{q}$. So this shows $f \in q$, as desired.


