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If $R=k[x_1,...,x_n]$ is a polynomial ring in $n$ indeterminates over a field $k$ of characteristic $0$, there is a characterization of the symbolic square of a prime ideal $P$ (the $n$-th symbolic power is defined as the $P^{(n)}:=P^nR_P\cap R$) due to Zariski and Nagata, which is as follows $$P^{(2)}=\{f\in P: \frac{\partial f}{\partial x_i}\in P, i=1,2,...,n\}$$

This is a special case of a general result on radical ideals (based on differential operators). I was wondering if anyone is aware of a simpler proof for the symbolic square case.

The containment is easy: for if $f\in P^{(2)}$, then there is a nonzerodivisor $g$ on $P$ such that $fg\in P^2$. So, $f\frac{\partial g}{\partial x_i} + g\frac{\partial f}{\partial x_i} \in P$, so $g\frac{\partial f}{\partial x_i} \in P$, so $\frac{\partial f}{\partial x_i}\in P$.

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You are surely supposing $P\neq 0$ ? – darij grinberg Dec 7 '10 at 23:22
And, generally, that $f$ has no constant term? Or $f\in R$ should be $f\in P$? – darij grinberg Dec 7 '10 at 23:25
@darji: Sorry that was a typo. It should have been $f\in P$ in the characterization. Thanks for pointing it out. – Timothy Wagner Dec 8 '10 at 0:14

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