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Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of homogeneous degree $d$ polynomials in three variables (up to scaling), where $\delta_d = \frac{d(d+3)}{2}$. Define $\mathcal{A}$ to be space of degree $d$ curves with a strict node at the point $[1,0,0]$, ie

$$ \mathcal{A} := \{ f \in \mathcal{D}: f([1,0,0]) =0, ~~ \nabla f |_{[1,0,0]} =0, \quad det \nabla^2 f ([1,0,0]) \neq 0 \}$$

Is it true that a generic element of $\mathcal{A}$ has finitely many singular points, provided $d$ is sufficiently large? It seems to me that a much stronger statement should be true, ie a generic element of $\mathcal{A}$ has only one singular point. Is there some reference for this result? It seems to me this is true, but I am not sure how to prove this rigorously.

By generic, I mean that the set of curves that has only finitely many nodes forms a dense open subset of $\mathcal{A}$.

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    $\begingroup$ What curve has infinitely many nodes? $\endgroup$
    – Will Sawin
    Jul 29, 2012 at 4:55
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    $\begingroup$ A double line for instance. x^2 =0 $\endgroup$
    – Ritwik
    Jul 29, 2012 at 5:17

2 Answers 2

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Ritwik: I believe we have discussed this before. Consider the blowing up at $[1,0,0]$, $$ \nu:X\to \mathbb{P}^2 $$ with exceptional divisor $E$. Now for the family $\overline{\mathcal{A}}$ of plane curves $C$ of degree $d$ having a singularity at $[1,0,0]$, consider the family of transform curves $\nu^*C - 2E$. By considering curves such as $x^{d-2}y^2$ and $x^{d-2}z^2$, you see that this family of curves on $X$ is base point free ($\textbf{edit}$: in a neighborhood of $E$; curves such as $y^d$ and $z^d$ show the linear system is base point free everywhere). Therefore, by Bertini's theorem, a general member is smooth on $X$. Since $\nu$ is an isomorphism away from $[1,0,0]$, it follows that a general member $C$ of $\overline{\mathcal{A}}$ is smooth away from $[1,0,0]$. $\textbf{Edit}$: Also the intersection number of these curves with $E$ is $2$, so a general member on $X$ will intersect $E$ in precisely $2$ distinct points, proving that a general member $C$ of $\overline{\mathcal{A}}$ has an ordinary double point at $[1,0,0]$.

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  • $\begingroup$ Would this argument also work if I changed the question slightly: instead of looking at the space of curves with a strict node at [1,0,0], I look at the space of curves with a strict A_k node at [1,0,0]. I want to know if a generic element of A (or preferably closure of A) has only one singular point. Assume d is sufficiently large. $\endgroup$
    – Ritwik
    Jul 29, 2012 at 13:43
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    $\begingroup$ @Ritwik: The method works so long as there are "enough" curves with that singularity so that the transform curves on the blowing up give a base point free linear system. This should work for $A_k$ singularities at a single singular point $[1,0,0]$ if $k<d$ using the four curves $x^{d-k}y^k$, $x^{d-k}z^k$, $y^d$ and $z^d$. However, the method would fail, for instance, if you considered degree $3$ curves with $A_1$ singularities at two distinct points. In fact this is very closely related to the Harbourne-Hirschowitz and Segre conjectures. I recommend you learn those. $\endgroup$ Jul 29, 2012 at 13:48
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Yes, a generic curve with one node at a fixed point of $\mathbb P^2$ over a field of characteristic $0$ has only that singularity. This is true in any degree $d$ at least $2$. By looking at a generic union of $d$ lines, two of which pass through a point $p$, we see that a generic element of the linear system of curves or degree $d$ with a singularity at $p$ has a node at $p$, and no other fixed points. The result follows from Bertini's Theorem for linear systems.

[Edit:] for a more elementary approach, it is easy to show that the curve with equation $xyz^{d-2} + x^d + y^d$ has no singularities outside of the obvious one.

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  • $\begingroup$ Regarding the second approach: I agree that the specific curve you have written down has only one singularity. Why does this imply that a GENERIC curve in my family, has only one singularity? $\endgroup$
    – Ritwik
    Jul 29, 2012 at 11:31
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    $\begingroup$ Well, if you have a family of curves and one of them is reduced with $n$ nodes, all the neighboring ones are reduced with at most $n$ nodes, and no other singularities. The proof is not completely trivial, though. $\endgroup$
    – Angelo
    Jul 29, 2012 at 12:58
  • $\begingroup$ Is there a reference for this fact? Secondly, it seems believable that an open neighborhood of the curve you have given has at most one node. But does this immediately imply that there is an open dense subset of curves with at most one node? $\endgroup$
    – Ritwik
    Jul 29, 2012 at 13:40

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