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Let $P_1,\ldots,P_9$ be $9$ different points in a cubic $C$. Let $S$ be a sextic passing through $P_1,\ldots,P_8$ with multiplicity at least $2$ and passing through $P_9$ with multiplicity at least $1$. I have read that in this case $S$ passes through $P_9$ with multiplicity at least $2$, but I do not really know why.

It looks like some kind of generalization of Cayley-Bacharach Theorem:

If a cubic passes through $8$ of the $9$ points of intersection of two cubics, then it passes through the ninth.

Nevertheless, it is claimed to be a consequence of Abel's Theorem:

Let $X$ be a compact Riemann surface. Let $D$ be a divisor of degree $0$ on $X$. Then $D$ is the divisor of a meromorphic function iff it is zero under the Abel-Jacobi map.

I have not been able to relate these results.

By the way, we are allowed to impose any condition on the points $P_1,\ldots,P_9$.

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  • $\begingroup$ Even in the case of Cayley-Bacharach, you can see that the 9 points are not arbitrary, they are the intersection of two cubics. In the sextic case, without any conditions on the 9 points, what you say will be wrong, but if they are the intersection of $C$ with another cubic, you will be alright. $\endgroup$ – Mohan Dec 5 '17 at 14:05
  • $\begingroup$ Dear @Mohan, I would appreciate if you could explain why. I am not able to prove why it suffices to take the points to be the intersection of two cubics. Thanks. $\endgroup$ – Zé Pequeno Dec 5 '17 at 16:57
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If the 9 points, $P_1,\ldots,P_9$ are the intersection of $C$ with another cubic $D$, then $2D$ gives you the divisor $\sum 2P_i$. If $S$ gives you $2P_1+\cdots +2P_8+P_9+P_{10}$, then these two divisors are linearly equivalent and thus $P_9\sim P_{10}$. On a cubic, this means $P_1=P_{10}$.

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  • $\begingroup$ Ok, thank you. If I am not wrong, what we need is the existence of a sextic giving the divisor $\Sigma 2P_i$. Could we take $P_1,\ldots,P_9$ such that this sextic exist but there is no cubic through them apart from $C$? $\endgroup$ – Zé Pequeno Dec 6 '17 at 10:06
  • $\begingroup$ Once the first eight points are specified, the ninth is uniquely determined, and there is a $1$-parameter family of cubics through those nine points. The most general solution to your problem seems to be this: If $P'_9$ is the point determined by $P_1, \dots, P_8$, then $2P_1+ \dots + 2P_8+P_9+P_{10}$ is linearly equivalent to $2P_1+ \dots + 2P_8+2P'_9$, so $P_9+P_{10}$ is linearly equivalent to $2P_9$. Then $P_9=P_{10}$ if and only if the tangent line at $P_9$ meets the curve in the same further point $Q$ as the tangent line at $P'_9$. There are three such $P'_9$ besides $P_9$ itself. $\endgroup$ – John Brevik Dec 6 '17 at 17:03

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