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Given a polynomial $f(x_1,... ,x_n)\in \mathbb{C}[x_1, ... ,x_n]$, we can formulate its (formal) partial derivative with respect to each of the $x_i$, say $f_{i}$. If $f\in \mathfrak{p}$ and $f_{i}\in \mathfrak{p}$ for each $i$, $\mathfrak{p}$ a prime ideal, do we have $f\in \mathfrak{p}^2$ in general?

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I might be missing something, but if $f(x)=x^2$, then $f_1(x)=2x\in(x)$ (this is with $n=1$) but $f_1(x)\notin (x)^2$. –  Alex Youcis Dec 13 '12 at 0:51
    
sorry for the serious typo. –  Ash GX Dec 13 '12 at 1:11

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Yes, this is true. (Edit: The comments below show that this argument only works for $\mathfrak p$ such that $\mathfrak p^2$ is primary. If this is not the case, the argument below gives that there exists $s\notin\mathfrak p$ such that $sf\in\mathfrak p^2.$ In particular, for $\operatorname{ord}_Z (f)$ we must localize at $Z.$)

In the case $\mathfrak p=\mathfrak m_a=(x_1-a_1,\ldots,x_n-a_n)$ is the maximal ideal of a point $a\in\Bbb A^n_{\Bbb C},$ the fact that $f\in\mathfrak m_a$ is equivalent to $f(a)=0.$ The fact that $\frac{\partial f}{\partial x_i}(a)=0$ for all $i=1,\ldots,n$ is equivalent to the linear term $f^{(1)}(x)=\sum_{i=1}^n\frac{\partial f}{\partial x_i}(a)(x_i-a_i)$ of the Taylor expansion $f(a)+f^{(1)}(x)+\cdots+f^{(m)}(x)$ of $f$ at $a$ being identically zero at $a.$ In other words $f(x)=f^{(2)}(x)+\cdots+f^{(m)}(x)$ at $a,$ which is equivalent to $f\in\mathfrak m_a^2.$

In the case of an arbitrary prime $\mathfrak p\in\Bbb C[x_1,\ldots,x_n],$ we know that $f\in\mathfrak p$ is equivalent to $f\in\mathfrak m_a$ for all maximal ideals $\mathfrak m_a\supseteq\mathfrak p.$ Geometrically, $f$ vanishes along the irreducible subvariety $V(\mathfrak p)$ if and only if $f$ vanishes at every (closed) point $a\in V(\mathfrak p).$ Suppose that $f\in\mathfrak p\subseteq\mathfrak m_a.$ By the previous, if $\frac{\partial f}{\partial x_i}\in\mathfrak p$ for every $i=1,\ldots,n,$ then $f\in\mathfrak m_a^2.$ Thus, $f\in\bigcap_{\mathfrak p\subseteq\mathfrak m_a}\mathfrak m_a^2.$

Let $Z=V(\mathfrak p)\subseteq\Bbb A^n_{\Bbb C},$ and let $I=(f)\subseteq\Bbb C[x_1,\ldots,x_n].$ The order of vanishing of $I$ along $Z,$ denoted $\operatorname{ord}_Z(I),$ is defined generally as

$$\max\{k~\vert~ I\subseteq I_Z^k\}=\max\{k~\vert~ (f)\subseteq \mathfrak p^k\}=\max\{k~\vert~ f\in\mathfrak p^k\},$$ and it is known that over algebraically closed fields this coincides with

$$\min_{a\in Z}\operatorname{ord}_a I=\min_{a\in Z}\left(\max\{k~\vert~ f\in\mathfrak m_a^k\}\right),$$ where $a\in Z$ denotes closed points. Since $f\in \mathfrak m_a^2,$ we have $\max\{k~\vert~ f\in\mathfrak m_a^k\}\ge 2$ for all $a\in Z,$ which in particular implies that $\operatorname{ord}_Z(I)\ge2,$ i.e., $f\in\mathfrak p^2.$

See Appendix A of this paper for more information about vanishing order of ideals.

(I initially wanted to show directly/algebraically that $f\in\bigcap_{\mathfrak p\subseteq\mathfrak m_a}\mathfrak m_a^2$ implies that $f\in\mathfrak p^2,$ but haven't quite gotten it to work yet. I'd be happy to see it done however, and will think a bit more on it.)

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It is not hard to see that the condition $f, f_i\in\mathfrak p$ is equivalent to there exists $s\notin \mathfrak p$ such that $sf\in\mathfrak p^2$. So a positive anwer to the OP is equivalent to saying that $\mathfrak p^2$ is a primary ideal. I wonder whether this is true over any field. The exercise in Hauser's paper you pointed to implies that over an algebraically closed field, $\mathfrak p^d$ is primary for any $d\ge 1$. –  user18119 Dec 13 '12 at 21:49
    
Very nice, thanks @QiL for pointing this out! –  Andrew Dec 14 '12 at 0:58
    
Dear Andrew, but I am not sure this is true that $\mathfrak p^2$ is primary. In fact, for an integral subvariety $Z=V(\mathfrak p)$, ord$_Z(f)$ is the order of $f$ at the generic point of $Z$, so one can't conclude that $f\in \mathfrak p^2$ (at least from Hauser's paper). –  user18119 Dec 14 '12 at 8:12
    
And my comment on the exercise is also wrong. –  user18119 Dec 14 '12 at 8:18
    
Dear @QiL, I see. I think my mistake was in reading Hauser's definition -- I didn't notice that we must take powers of the ideals in the local ring of $Z$! Thus, what my argument actually shows is what you mention, there exists $s\notin\mathfrak p$ such that $sf\in\mathfrak p^2.$ It seems that Hauser's exercise gives the analogous result for higher order, i.e., that $f,f_i,f_{ij},\ldots,f_{i_1\ldots i_d}\in\mathfrak p$ is equivalent to $\exists s\notin\mathfrak p$ such that $sf\in\mathfrak p^d.$ –  Andrew Dec 14 '12 at 14:41

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