Please help me to solve the following problem:

There are curves $C$ of degree $3$ and $Q$ of degree $2$ in $\mathbb{C}P^2$. $P_1, \ldots,P_6$ are intersection points of $C$ and $Q$. $T_i$ is a tagent to $C$ at $P_i$, $i \in \{1,\ldots, 6\}$. $Q_i$ is point of intersection of $T_i$ and $C$.

I need to prove that $Q_1, \ldots,Q_6$ are lying on another curve $Q'$ of degree $2$.

My idea:

I think (I am not 100% sure) that I should somehow use: Cayley–Bacharach theorem. Can you please advice how to use it?

Thanks a lot for your help!

  • $\begingroup$ What is the source of this question ? $\endgroup$ – Rene Schipperus Mar 9 '18 at 22:34
  • $\begingroup$ I am self-studding book Algebraic Curves by R.J.Walker, this is simplified version of ex. 7.3.5 $\endgroup$ – Hedgehog Mar 9 '18 at 22:35
  • $\begingroup$ Which chapter, page ? $\endgroup$ – Rene Schipperus Mar 9 '18 at 22:39
  • $\begingroup$ Chapter 4, paragraph 7, - Exercise 5 simplified by myself. In my translated version it is page 144, but I have no idea about original page number. $\endgroup$ – Hedgehog Mar 9 '18 at 22:42
  • $\begingroup$ To my knowledge a conic and quadrics are both degree two curves and can not meet in six points. What is your definition? $\endgroup$ – Mohan Mar 9 '18 at 22:48

This is a straight forward copy of the proof of Thm7.3. Let $D$ be a conic through five of the six points $Q_i$ and $L$ any line through the sixth, not passing though the other points. Let $G$ be the product of the lines $T_i$.

Then by Noether $$Q^2DL=CA+GB$$ now if $r$ and $s$ are the other two points of intersection of $L$ and $C$ then they must be zeros of $GB$ and since they are not zeros of $G$, by choice of $L$ we have that they are zeros of $B$. However $B$ is linear so $B=L$, and we have $$Q^2DL=CA+GL$$ so that gives that $Q^2D$ and $G$ have exactly the same intersection with $C$ and thus $D$ must go through all the points $Q_i$.

  • $\begingroup$ Great, thanks a lot! $\endgroup$ – Hedgehog Mar 10 '18 at 4:56

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