0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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, - 4.7.3.5. 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
1
$\begingroup$

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 $Q_i$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.