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Let $X\subseteq \mathbb{P}^n$ be a closed irreducible subvariety, with vanishing ideal $I(X)\subseteq k[x_0,\ldots,x_n]$, where $k$ is the ground field, assumed to be algebraically closed. Let $F\in k[x_0,\ldots,x_n]$ be a homogeneous polynomial. If $F$ belongs to the ideal $I(X)^2$, then $F$ is singular along $X$ (the singular locus of $F$ contains $X$). Is the converse true?

I am mostly interested in the case where $n=3$ and $X$ is a curve, of small degree (say $\le 9$). But the general case seems interesting too.

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  • $\begingroup$ When you say $F$ is singular on $X$, do you mean that the singular locus of $F$ contains $X$? $\endgroup$
    – Mohan
    Commented Jun 28, 2023 at 14:45
  • $\begingroup$ Yes, this is exactly what I mean. I edited the question accordingly. $\endgroup$
    – Jérémy Blanc
    Commented Jun 28, 2023 at 14:48

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If $X=\mathbb{V}(I)$ is given by the ideal $I$, then the $m$th symbolic power $I^{[m]}$ consists of all those functions vanishing to multiplicity $m$ at the generic point of $X$. Thus a hypersurface $\mathbb{V}(F)$ will be singular along $X$ if and only if $F\in I^{[2]}$. (Thanks to Zach Teitler for pointing this out in the comments below.) To give a counterexample it is therefore enough to find an $X$ for which $I^2\subsetneq I^{[2]}$.

A (non-irreducible, affine) example is $X=\mathbb{V}(xy,yz,zx)$, the union of the three coordinate axes in $\mathbb{A}^3$ and $F=xyz\notin I^2$. Then the hypersurface $\mathbb{V}(xyz)$ has multiplicity $2$ on every component of $X$. In general if $X\subset \mathbb{P}^3$ is an irreducible space curve which is smooth apart from a triple point (i.e. a point locally analytically isomorphic to the example above) then there should also be function $F\in I^{[2]}\setminus I^2$ as above.

(P.S. If you want to compute $I^{[m]}$ explicitly (e.g. in the case you are interested in, of a curve $X$ of small degree) then you can ask your favourite computer algebra package for the primary decomposition of $I^m$. The unique primary component supported on the whole of $X$ will be $I^{[m]}$.)

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    $\begingroup$ Sorry, what do you mean by saying that looking at $I^{[2]}$ isn't enough? All the singular hypersurfaces are in there (and it contains the higher symbolic powers anyway). Of course you need to look at higher symbolic powers if you're interested in resurgence or Waldschmidt constants or symbolic Rees algebras or other things. But I think for the original question the answer is just $I^{[2]}$. $\endgroup$
    – Zach Teitler
    Commented Jun 29, 2023 at 3:11
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    $\begingroup$ Sorry, you are quite right, it is is clear from the definition that $I^{[m]}\subset I^{[2]}$ for all $m\geq3$. (I had mistakenly thought that the theorem I quoted might imply the existence of an $I$ such that $I^2=I^{[2]}$, but with some $f\in I^{[k]}\setminus I^2$ for some $3\leq k < 2n$.) $\endgroup$
    – Tom Ducat
    Commented Jun 29, 2023 at 9:24
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    $\begingroup$ Thanks for your answer. It is a nice one. $\endgroup$
    – Jérémy Blanc
    Commented Jun 29, 2023 at 13:55

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