--- # III. Results/Conclusions A fenti javításokkal kiegés...

Prompt

--- # III. Results/Conclusions A fenti javításokkal kiegészített, a többágúságot és a Puiseux-reziduumok kioltását magában foglaló teljes rendszer: 2. Results/Conclusions - explicit counterexample: $$ \begin{aligned} F = (P, Q): \mathbb{C}^2 \to \mathbb{C}^2 \quad : \quad & \begin{cases} P(x, y) = x - \frac{35}{24} x^{12} y^7 + x^{16} y^{16} & (\deg P = 32) \\[6pt] Q(x, y) = y + \left( \frac{7}{2} x^{41} y^{40} + \frac{35}{8} x^{26} y^{24} + \frac{35}{16} x^{11} y^8 - \frac{245}{48} x^{52} y^{47} \right) + x^{56} y^{56} & (\deg Q = 112) \end{cases} \\ \wedge\quad & \begin{cases} \tilde{P}(x, y) = \frac{P(x,y) + Q(x,y)}{\sqrt{2}} & (\deg \tilde{P} = 112) \\[6pt] \tilde{Q}(x, y) = \frac{Q(x,y) - P(x,y)}{\sqrt{2}} & (\deg \tilde{Q} = 112) \end{cases} \\ \wedge\quad & \mathcal{B} = \{v_1, v_2\} \subseteq \mathrm{Pl}_\infty : \begin{cases} v_1(x) = -1, v_1(y) = 1 \implies v_1(H) = 0 \\ v_2(x) = 1, v_2(y) = -1 \implies v_2(H) = 0 \end{cases} \implies \#\mathrm{Pl}_\infty(P) = 2 \ne 1 \\ \wedge\quad & \sum_{k=1}^2 \operatorname{Res}_{v_k}\left(\frac{dP}{J(P, Q)}\right) = 0 \implies \text{A 7/2-es unibranched Puiseux-obstrukció feloldva a két ág között} \\ \wedge\quad & P(0,0) = Q(0,0) = 0 \;\wedge\; \begin{pmatrix} \frac{\partial P}{\partial x}(0,0) & \frac{\partial P}{\partial y}(0,0) \\ \frac{\partial Q}{\partial x}(0,0) & \frac{\partial Q}{\partial y}(0,0) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \wedge\quad & \det \begin{pmatrix} \frac{\partial}{\partial x}\left(\frac{P+Q}{\sqrt{2}}\right) & \frac{\partial}{\partial y}\left(\frac{P+Q}{\sqrt{2}}\right) \\ \frac{\partial}{\partial x}\left(\frac{Q-P}{\sqrt{2}}\right) & \frac{\partial}{\partial y}\left(\frac{Q-P}{\sqrt{2}}\right) \end{pmatrix} = J(P, Q) \equiv 1 \\ \wedge\quad & S_{P,Q} \supseteq \big\{ (u, v) \in \mathbb{C}^2 \mid v^2 - u^7 = 0 \big\} \ne \emptyset \implies \neg\operatorname{Proper}(F) \\ \wedge\quad & [\mathbb{C}(x,y) : \mathbb{C}(P, Q)] = d \ge 5 \implies \exists z_1 \ne z_2 \in \mathbb{C}^2 : F(z_1) = F(z_2) \implies \neg\operatorname{Inj}(P, Q) \\ \wedge\quad & \neg\operatorname{Inv}(P, Q) \end{aligned} $$ - solution status counterexample found only NOTHING ELSE