Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|cite|improve this question

3 Answers 3

up vote 9 down vote accepted

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.

share|cite|improve this answer
Thank you very much! – Semen Podkorytov Nov 12 '10 at 10:22

The answer "yes" follows immediately from Corollary 2 in "A Nullstellensatz with Nilpotents and Zariski’s Main Lemma on Holomorphic Functions" by D. Eisenbud and M. Hochster, J. Algebra 58 (1979), 157-161,

share|cite|improve this answer

I think that the answer is 'yes'. You should estimate how 'far' I is from its nilradical (for which one can take $N=0$). Since the ring of polynomials is Noetherian, you could obtain the nilradical of I by adding a finite number of roots of elements of I to it (i.e. you obtain some $I_1$, add some root of its element, get $I_2$, etc.). I strongly suspect that you can get some value of $N$ looking at this process.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.