answer:a) To find the rational points on the elliptic curve y^2 = x^3 + 4x, we can use the method of descent. First, we note that the point (0,0) is a rational point on the curve. Next, we consider the line y = mx + b that passes through (0,0) and another rational point (x', y') on the curve. We have: y^2 = (mx + b)^2 = x^3 + 4x Substituting y = mx + b and x = 0, we get b = 0. So, the line is y = mx. Now, we substitute y = mx in the elliptic curve equation: (m^2)x^2 = x^3 + 4x Rearranging the equation, we get: x^3 - m^2x^2 + 4x = 0 Factoring out x, we have: x(x^2 - m^2x + 4) = 0 Since x = 0 is already a rational point, we focus on the quadratic factor: x^2 - m^2x + 4 = 0 Using the quadratic formula, we get: x = (m^2 ± √(m^4 - 16))/2 For x to be rational, the discriminant (m^4 - 16) must be a perfect square. Let m^4 - 16 = n^2 for some rational number n. Then, we have: m^4 - n^2 = 16 This is a difference of squares, so we can factor it as: (m^2 - n)(m^2 + n) = 16 Now, we consider the divisors of 16: (1, 16), (2, 8), and (4, 4). Solving these systems of linear equations, we find the following rational values for (m, n): (1) m^2 - n = 1, m^2 + n = 16: m^2 = 8.5, which is not rational. (2) m^2 - n = 2, m^2 + n = 8: m^2 = 5, n = 3, m = ±√5 (3) m^2 - n = 4, m^2 + n = 4: m^2 = 4, n = 0, which is not possible. So, the only possible rational value for m is ±√5. Substituting this value into the equation for x, we get: x = (5 ± √(5^4 - 16))/2 = (5 ± 3)/2 This gives us two rational values for x: x = 1 and x = 4. Substituting these values back into the equation y = mx, we find the corresponding rational points on the elliptic curve: For x = 1: y = ±√5 For x = 4: y = ±4√5 Thus, the rational points on the elliptic curve are (0, 0), (1, √5), (1, -√5), (4, 4√5), and (4, -4√5). b) To find a birational transformation that maps the elliptic curve to a circle, we can use the Weierstrass normal form. The given elliptic curve is already in the form y^2 = x^3 + Ax + B, with A = 0 and B = 4. The Weierstrass normal form is: y^2 = x^3 + Ax + B = (x^2 + u^2)(x^2 + v^2) We want to find a rational transformation (x, y) -> (u, v) such that the equation holds. We can try the following transformation: u = x/y v = y/x Solving for x and y, we get: x = uv/(u^2 + v^2) y = (u^2 - v^2)/(u^2 + v^2) Substituting these expressions into the elliptic curve equation, we get: ((u^2 - v^2)^2)/(u^2 + v^2)^2 = (u^2v^2)/(u^2 + v^2)^2 + 4(uv/(u^2 + v^2)) Simplifying, we get: (u^2 - v^2)^2 = u^2v^2 + 4(u^2 + v^2)(uv)^2 This equation represents a circle in the (u, v)-plane. Thus, the birational transformation (x, y) -> (u, v) with x = uv/(u^2 + v^2) and y = (u^2 - v^2)/(u^2 + v^2) maps the elliptic curve to a circle.

answer:To show that C is birational to the affine line mathbb{A}^1, we need to find a rational map between them. Let's first find the parametrization of the curve C. Since C is defined by the equations x=y and z=0, we can substitute these into the equation of the algebraic variety: x^2 + x^2 + 0^2 - 3 = 0 Rightarrow 2x^2 = 3 Rightarrow x^2 = frac{3}{2}. Now, let t = x = y. Then, t^2 = frac{3}{2}, and the curve C can be parametrized as (t, t, 0), where t^2 = frac{3}{2}. Now, let's find a birational map from C to mathbb{A}^1. We can define the map phi: C dashrightarrow mathbb{A}^1 as follows: phi(t, t, 0) = t. This map is well-defined since t is uniquely determined by the point (t, t, 0) on the curve C. Moreover, it is a rational map since it is given by a rational function of the coordinates. Now, let's find the inverse map phi^{-1}: mathbb{A}^1 dashrightarrow C. Given a point t in mathbb{A}^1, we can define the inverse map as follows: phi^{-1}(t) = (t, t, 0). This map is well-defined since the point (t, t, 0) on the curve C is uniquely determined by the parameter t. Moreover, it is a rational map since it is given by a rational function of the coordinates. Thus, we have found an explicit birational map between the curve C and the affine line mathbb{A}^1. The curve C is birational to the affine line mathbb{A}^1.

answer:Since X is birational to the projective plane P^2, there exists a rational map ϕ : X → P^2 and its inverse ψ : P^2 → X such that ϕ ◦ ψ and ψ ◦ ϕ are the identity maps on P^2 and X, respectively. To find the rational map ϕ, we need to find a set of rational functions on X that separates points and tangent vectors. Let f(x, y, z) be a homogeneous polynomial of degree 3. We can write it as: f(x, y, z) = a_0x^3 + a_1x^2y + a_2xy^2 + a_3y^3 + a_4x^2z + a_5xyz + a_6y^2z + a_7xz^2 + a_8yz^2 + a_9z^3 = 0 Now, consider the following rational functions on X: g_1(x, y, z) = x/z g_2(x, y, z) = y/z These functions are well-defined on X, except at the point [0:0:1]. We can define the rational map ϕ : X → P^2 as: ϕ([x:y:z]) = [g_1(x, y, z) : g_2(x, y, z) : 1] = [x:y:z] This map is well-defined on X, except at the point [0:0:1]. However, since X is birational to P^2, there exists an open subset U ⊆ X such that ϕ is an isomorphism between U and its image ϕ(U) ⊆ P^2. Thus, we can find the inverse map ψ : P^2 → X by restricting ϕ to U and taking its inverse. To summarize, the rational map ϕ : X → P^2 induced by the birational morphism is given by: ϕ([x:y:z]) = [x:y:z]

answer:To find the intersection points of C and L in the projective plane P^2, we need to substitute the equation of L (x = 0) into the equation of C and solve for the homogeneous coordinates [x : y : z]. Substituting x = 0 into the equation of C, we get: 0^3 + y^3 - 3(0)y = y^3 = 0 Since we are in the projective plane, we cannot simply say y = 0, as this would lead to the point [0 : 0 : z] which is not a valid point in P^2. Instead, we need to consider the curve in homogeneous coordinates. The equation of C in homogeneous coordinates is: X^3 + Y^3 - 3XY*Z = 0 Now, we substitute X = 0 (the homogeneous version of x = 0) into the equation: 0^3 + Y^3 - 3(0)Y*Z = Y^3 = 0 Since Y^3 = 0, we have Y = 0. Therefore, the intersection points have the form [0 : 0 : Z]. However, as mentioned earlier, [0 : 0 : Z] is not a valid point in P^2. To find a valid point, we need to look at the line at infinity, which is given by Z = 0. So, we substitute Z = 0 into the equation of C in homogeneous coordinates: X^3 + Y^3 - 3XY*0 = X^3 + Y^3 = 0 Since X^3 + Y^3 = 0, we have X^3 = -Y^3. This implies that X = Y (up to a scalar multiple, which is allowed in projective coordinates). Therefore, the intersection points have the form [X : X : 0]. Thus, the intersection points of C and L in the projective plane P^2 are [1 : 1 : 0] and all other points of the form [X : X : 0] where X ≠ 0.