answer:To calculate the degree of K_X + D, we first need to find the canonical divisor K_X. For a nonsingular projective plane curve given by the equation F(x, y, z) = 0, the canonical divisor is given by K_X = (d-3)H, where d is the degree of the curve and H is a hyperplane section. In our case, the curve X is given by the equation y^2z = x(x-z)(x-lambda z). The degree of this curve is the highest total degree of any monomial term, which is 3 (from the term y^2z). Therefore, d = 3. Now we can find the canonical divisor K_X = (d-3)H = (3-3)H = 0. Since we are given that deg(D) = 3, we can now calculate the degree of K_X + D: deg(K_X + D) = deg(K_X) + deg(D) = 0 + 3 = 3.

answer:To determine the degree and dimension of the linear series |2P|, we first need to compute the divisor 2P. Since P = [1, 1] is a point on the curve C, we have: 2P = P + P To compute P + P, we need to find the tangent line to the curve C at the point P. To do this, we first compute the partial derivatives of the curve equation with respect to x and y: ∂C/∂x = 5x^4 + 1 ∂C/∂y = 2y Now, we evaluate these partial derivatives at the point P = [1, 1]: ∂C/∂x(P) = 5(1)^4 + 1 = 6 ∂C/∂y(P) = 2(1) = 2 The tangent line to the curve C at the point P is given by the equation: y - y(P) = m(x - x(P)) where m is the slope of the tangent line, which is given by the ratio of the partial derivatives: m = - (∂C/∂x(P)) / (∂C/∂y(P)) = -6/2 = -3 So, the tangent line equation is: y - 1 = -3(x - 1) Now, we substitute this equation into the curve equation to find the intersection points: (1 - 3(x - 1))^2 = x^5 + x + 1 Expanding and simplifying, we get: x^5 - 8x^4 + 21x^3 - 22x^2 + 9x = 0 Since P is a point on the tangent line, it must be a double root of this polynomial. By factoring out (x - 1)^2, we get: (x - 1)^2(x^3 - 6x^2 + 13x - 9) = 0 The remaining cubic factor corresponds to the other intersection point Q of the tangent line with the curve C. To find the coordinates of Q, we need to solve the cubic equation: x^3 - 6x^2 + 13x - 9 = 0 This cubic equation has one real root, which can be found numerically (approximately x ≈ 2.879). We can then substitute this value of x back into the tangent line equation to find the corresponding y-coordinate (approximately y ≈ -4.637). Now, we have the coordinates of the point Q, and since 2P = P + P = Q, the divisor 2P is given by: 2P = Q ≈ [2.879, -4.637] The degree of the divisor 2P is the sum of the coefficients of the points in the divisor, which is 2 in this case. The dimension of the linear series |2P| is given by the Riemann-Roch theorem: l(2P) = g - l(K - 2P) + deg(2P) + 1 where g is the genus of the curve (g = 2), K is the canonical divisor, and l(K - 2P) is the dimension of the linear series |K - 2P|. Since the curve is of genus 2, the canonical divisor K has degree 2g - 2 = 2. The degree of the divisor 2P is 2, so the degree of the divisor K - 2P is 0. This means that l(K - 2P) = 1. Substituting these values into the Riemann-Roch theorem, we get: l(2P) = 2 - 1 + 2 + 1 = 4 So, the dimension of the linear series |2P| is 4. In conclusion, the degree of the linear series |2P| is 2, and the dimension is 4.

answer:Since C is an elliptic curve, we have g = 1. The divisor D has degree d = mathrm{deg}(P - Q) = mathrm{deg}(P) - mathrm{deg}(Q) = 1 - 1 = 0. Now we can apply the Riemann-Roch theorem: mathrm{dim}(H^0(C,O_C(D))) = d - g + 1 + mathrm{dim}(H^1(C,O_C(D))) = 0 - 1 + 1 + mathrm{dim}(H^1(C,O_C(D))) = mathrm{dim}(H^1(C,O_C(D))) Since D has degree 0, we know that O_C(D) is a line bundle of degree 0. For an elliptic curve, the degree 0 line bundles are precisely the ones with a non-vanishing global section, so we have: mathrm{dim}(H^0(C,O_C(D))) = 1 Thus, the dimension of the space of global sections of O_C(D) is 1.

answer:Let A = (x_1, y_1, z_1) and B = (x_2, y_2, z_2) be the homogeneous coordinates of points A and B on the curve C. Since A and B are on C, we have: f(x_1, y_1, z_1) = x_1^4 + y_1^4 + z_1^4 = 0 f(x_2, y_2, z_2) = x_2^4 + y_2^4 + z_2^4 = 0 Now, consider a degree-2 divisor (conic) passing through both A and B. The equation of such a conic can be written as: g(x, y, z) = a_1x^2 + a_2y^2 + a_3z^2 + a_4xy + a_5xz + a_6yz Since A and B are on this conic, we have: g(x_1, y_1, z_1) = a_1x_1^2 + a_2y_1^2 + a_3z_1^2 + a_4x_1y_1 + a_5x_1z_1 + a_6y_1z_1 = 0 g(x_2, y_2, z_2) = a_1x_2^2 + a_2y_2^2 + a_3z_2^2 + a_4x_2y_2 + a_5x_2z_2 + a_6y_2z_2 = 0 These are two linear equations in the six unknowns a_1, a_2, a_3, a_4, a_5, a_6. Since we are looking for a conic that is also a divisor on C, we need to find the intersection of the conic g(x, y, z) = 0 with the curve f(x, y, z) = 0. This intersection can be computed using the resultant of the two polynomials f and g with respect to one of the variables, say z. The resultant will be a polynomial in x and y with coefficients depending on the unknowns a_1, a_2, a_3, a_4, a_5, a_6. Since C is a smooth curve of degree 4, the intersection of C with a conic should consist of 8 points (by Bézout's theorem). However, we already know that two of these points are A and B. Therefore, the resultant must be a polynomial of degree 6 in x and y. Now, we have a system of linear equations involving the coefficients of the resultant and the unknowns a_1, a_2, a_3, a_4, a_5, a_6. Solving this system, we can find the number of linearly independent solutions, which corresponds to the number of distinct degree-2 divisors on C that pass through both A and B. The system has 6 unknowns and 6 equations (2 from the conditions that A and B are on the conic, and 4 from the conditions on the coefficients of the resultant). Since the curve C is smooth, the system has a unique solution, which corresponds to a single degree-2 divisor on C that passes through both A and B. Therefore, there is only 1 degree-2 divisor on C that passes through both A and B.