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Let $x_1,...,x_7$ be distinct points in $\mathbb{C}^2$. Prove that there exists a cubic curve passing through these points which has a singularity at the point $x_1$.

My attempt so far... A related question was: show that for every five points $a_1,...,a_5\in\mathbb{P}^2$, there is a conic containing them. In the given solution it was (roughly) argued, that the space of quadratic forms is a $6$-dimensional vector space and that the condition $q(a_i)=0$ for the conic $q=0$ is a linear equation in the coefficients of $q$. Now, imposing $5$ linear conditions on a $6$-dimensional vector space leaves at least a $1$-dimensional space of solutions - which is the desired conic.

Is there a way to apply this approach to the given question? If so, I assume the space of cubic forms has dimension $10$, is this correct? I don't know how the singularity would show up here.

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  • $\begingroup$ Hi @casimir could you give me the title of the books thats you are working on ? $\endgroup$ – Bernstein Jul 23 '18 at 16:03
  • $\begingroup$ I'm not working with a book, only with lecture notes ... and they don't go into any detail with respect to the question asked. $\endgroup$ – casimir Jul 23 '18 at 16:05
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    $\begingroup$ thank you very much , good luck with your question $\endgroup$ – Bernstein Jul 23 '18 at 16:13
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    $\begingroup$ Sottile discusses the $12$ nodal cubics through $8$ points. $\endgroup$ – Jan-Magnus Økland Jul 23 '18 at 16:35
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    $\begingroup$ This is indeed quite similar to the conic question. Can you see that there exists a cubic curve passing through $x_1,\ldots,x_7$? (This corresponds to 7 equations on your coefficients). Asking for your curve to have a singularity at $x_1$ imposes some more equations (how many?). $\endgroup$ – loch Jul 23 '18 at 17:17
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Let $\mathbb{P}^N\equiv\mathbb{P}(\mathbb{C}[z_0,z_1,z_2]_3)$ be the space of plane cubic (projective) curves; its dimension $N$ is \begin{equation} \binom{3-1+3}{3}-1=10-1=9. \end{equation} Let $\displaystyle F=\sum_{i_0+i_1+i_2=3}a_{i_0i_1i_2}z_0^{i_0}z_1^{i_1}z_2^{i^2}$ be the generic homogeneous polynomial of degree $3$ in three variables; let \begin{equation} \nu_{2,3}:\mathbb{P}^2\to\mathbb{P}^9 \end{equation} be the Veronese embedding of the (complex) projective plane in $\mathbb{P}^9$; one knows that the image of $F$ via $\nu_{2,3}$ is a hyperplane $H$.

The conditions $F([1:x_k])=0$, where $k\in\{1,\dots,7\}$ and $[1:x_k]$ are the homogeneous cordinates of $x_k$'s in $\mathbb{P}^2$, constraint $H$ passing through $7$ distinct points of $\mathbb{P}^9$: this is possible, so there exist cubic plane curves passing through the points $x_k$.

Remark 1. If one has $9$ distinct points in $\mathbb{P}^2$, then the hyperplane $H$ is uniquely determined; in other words, on $9$ distinct points in $\mathbb{P}^2$ gets through a unique cubic plane curve.

Let $F_{z_1}\equiv F_1$ and $F_{z_2}\equiv F_2$ be the partial derivaties of $F$; the curve $\Gamma=\{F(1:z_1:z_2)=0\}$ has a (unique) singular point if the system \begin{equation} \begin{cases} F(1:z_1:z_2)=0\\ F_1(1:z_1:z_2)=0\\ F_2(1:z_1:z_2)=0 \end{cases} \end{equation} admits a (unique) solution; by the previous reasoning, one fixes $7$ of $10$ coefficients of $F$, so the previous system can have at most a unique solution after fixed the other $3$ coefficients of $F$.

Finded this solution, one has determined a cubic plane curve $\Gamma$ passing through the points $x_k$ with a singular point.

Remark 2. The projective closure of $\Gamma$ is an elliptic curve.

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  • $\begingroup$ Very helpful, thank you! $\endgroup$ – casimir Jul 27 '18 at 9:57
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    $\begingroup$ You are welcome. P.S.: The trick of Veronese embedding works also for the generic (algebraic) hypersurface of degree $d\geq1$. ;) $\endgroup$ – Armando j18eos Jul 27 '18 at 10:01

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