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Sep 27, 2012 at 22:52 comment added Renato There is a typo on the equation. I am rephrasing it in another question. thanks for the help/interest
Sep 27, 2012 at 22:41 comment added Renato From Floer theory we have that if $(z,w)∈critW$ then $W^3(z,w)=27$. Solving for z in one of the partial derivatives of W, we get that this condition is equivalent to the roots of $f_1(w)$ being also roots of $f_2(w)$.
Sep 27, 2012 at 22:21 comment added Renato Sorry, rest = reminder , as Peter said "the remainder of $f_2(w)^10 $ modulo $f_1(w)$ vanishes". The origin of this question comes from Mirror Symmetry and computing the superpotential $W$, which is just function, in my case from $C^2$ to $C$, of the Landau-Ginzburg model for the complement of a divisor in $CP^2$. The Superpotential is a Laurent polynomial whose coefficients are given by a count of holomorphic discs, and monomials are related to a relative homotopy class of the group. I was able to prove that $W = z + \frac{2(1+w)^2}{z^2} + \frac{P_5(w)}{w z^5}$ –
Sep 27, 2012 at 9:59 comment added Peter Mueller Apparently Renato doesn't consider multiplicities. As $f_1(w)$ has degree at most $10$, his condition then is that the remainder of $f_2(w)^{10}$ modulo $f_1(w)$ vanishes. I believe `set the rest to zero' means the condition coming from setting the coefficients of this remainder to $0$. By the way, a solution to the question is $P(w)=w((w+1)^4+c)$ for any $c$. Also, $P(w)=-w(w+1)^4$ is a solution. Still, it would be nice to learn more about the background of this question!
Sep 26, 2012 at 23:36 comment added Igor Rivin What do you mean by "set the rest to zero"?
Sep 26, 2012 at 22:52 comment added Renato The idea is just to divide the second polynomial f_2(w) = (24)3(1+w)5w4[4P5(w)w−Q5(w)(1−w)]3+108(Q5(w))5 , to the tenth power by f_1 (w) = (Q5(w))2+16Q5(w)(1+w)3w2−80(1+w)2w3P5(w) then set the rest to zero. We should get up to nine conditions on c_i's. Using they are integers, I hope to get strong restrictions, or in the best case, one solution that is expected to be (1, 5 ,10 , 10 , 5, 1).
Sep 26, 2012 at 22:46 vote accept Renato
Sep 26, 2012 at 22:46
Sep 25, 2012 at 20:49 history answered Igor Rivin CC BY-SA 3.0