answer:To determine the degree of X as a projective variety in P^2, we first need to homogenize the polynomial f(x). We introduce a new variable, say z, and homogenize f(x) to a homogeneous polynomial F(x, y, z) in three variables x, y, and z. To homogenize f(x), we replace each term of f(x) with its homogeneous counterpart by adding powers of z such that the total degree of each term is equal to the degree of the polynomial, which is 4 in this case. So, the homogeneous counterpart of f(x) is: F(x, y, z) = x^4 - 5x^3z + 5x^2z^2 + 5xz^3 - z^4 Now, X is the algebraic set given by the zero locus of F(x, y, z) in P^2. The degree of X as a projective variety in P^2 is equal to the degree of the homogeneous polynomial F(x, y, z), which is the highest total degree of its terms. In this case, the degree of F(x, y, z) is 4, as the highest total degree of its terms is 4 (e.g., x^4, 5x^2z^2, and -z^4). Therefore, the degree of X as a projective variety in P^2 is 4.

answer:To prove that |mL-K_X-H| is base-point-free for all sufficiently large integer m, we will use the Kawamata-Viehweg vanishing theorem. First, recall that X is a Calabi-Yau variety of dimension 3, so it has a trivial canonical divisor, i.e., K_X ≅ O_X. Also, L is a very ample line bundle on X, and H is an ample divisor on X. Now, let's consider the line bundle mL-K_X-H. Since K_X ≅ O_X, we have: mL-K_X-H ≅ mL-H. We want to show that |mL-H| is base-point-free for all sufficiently large integer m. By the Kawamata-Viehweg vanishing theorem, we have: R^i(X, mL-H) = 0 for all i > 0 and for all sufficiently large integer m. In particular, for i = 1, we have: R^1(X, mL-H) = 0 for all sufficiently large integer m. Now, consider the short exact sequence of sheaves on X: 0 → O_X(-H) → O_X → O_H → 0. Tensoring this sequence with the line bundle mL, we get: 0 → mL(-H) → mL → (mL)|_H → 0. Applying the cohomology functor, we obtain the following long exact sequence: 0 → H^0(X, mL(-H)) → H^0(X, mL) → H^0(H, (mL)|_H) → H^1(X, mL(-H)) → ... By the Kawamata-Viehweg vanishing theorem, we know that H^1(X, mL(-H)) = 0 for all sufficiently large integer m. Therefore, the map: H^0(X, mL) → H^0(H, (mL)|_H) is surjective for all sufficiently large integer m. This implies that the linear system |mL-H| is base-point-free for all sufficiently large integer m. Now, let's show that |3L-K_X-H| is base-point-free. Since K_X ≅ O_X, we have: 3L-K_X-H ≅ 3L-H. By the argument above, we know that |mL-H| is base-point-free for all sufficiently large integer m. In particular, if m = 3 is sufficiently large, then |3L-H| is base-point-free.

answer:a) To show that X is a smooth projective variety, we need to check that the polynomial f(x, y, z, w) defines a smooth hypersurface in the projective space P^3. This means that the partial derivatives of f with respect to x, y, z, and w must not simultaneously vanish at any point on X. The partial derivatives of f are: ∂f/∂x = 4x^3 - 4xy^2 - 4xz^2 - 4xw^2 ∂f/∂y = 4y^3 - 4yx^2 - 4yz^2 - 4yw^2 ∂f/∂z = 4z^3 - 4zx^2 - 4zy^2 - 4zw^2 ∂f/∂w = 4w^3 - 4wx^2 - 4wy^2 - 4wz^2 Now, suppose that there exists a point (x₀, y₀, z₀, w₀) on X such that all these partial derivatives vanish simultaneously. Then we have: 4x₀^3 - 4x₀y₀^2 - 4x₀z₀^2 - 4x₀w₀^2 = 0 4y₀^3 - 4y₀x₀^2 - 4y₀z₀^2 - 4y₀w₀^2 = 0 4z₀^3 - 4z₀x₀^2 - 4z₀y₀^2 - 4z₀w₀^2 = 0 4w₀^3 - 4w₀x₀^2 - 4w₀y₀^2 - 4w₀z₀^2 = 0 Dividing each equation by 4, we get: x₀(x₀^2 - y₀^2 - z₀^2 - w₀^2) = 0 y₀(y₀^2 - x₀^2 - z₀^2 - w₀^2) = 0 z₀(z₀^2 - x₀^2 - y₀^2 - w₀^2) = 0 w₀(w₀^2 - x₀^2 - y₀^2 - z₀^2) = 0 If x₀ = 0, then the first equation implies y₀^2 = z₀^2 + w₀^2. Similarly, if y₀ = 0, then the second equation implies x₀^2 = z₀^2 + w₀^2. If both x₀ and y₀ are nonzero, then we can divide the first equation by x₀ and the second equation by y₀ to get: x₀^2 - y₀^2 - z₀^2 - w₀^2 = 0 y₀^2 - x₀^2 - z₀^2 - w₀^2 = 0 Adding these two equations, we get 2(x₀^2 - y₀^2) = 0, which implies x₀^2 = y₀^2. But then, the first equation becomes x₀^2 = z₀^2 + w₀^2, which is a contradiction since x₀ ≠ 0 and y₀ ≠ 0. Therefore, x₀ = y₀ = 0. Similarly, we can show that z₀ = w₀ = 0. But this is a contradiction since (0, 0, 0, 0) is not a point in the projective space P^3. Hence, there is no point on X where all the partial derivatives simultaneously vanish, and X is a smooth projective variety. b) To compute the Hodge diamond of X, we can use the Lefschetz hyperplane theorem, which states that for a smooth hypersurface X in P^n of degree d, we have: h^{p, q}(X) = h^{p, q}(P^n) for p + q < n - 1 h^{p, q}(X) = h^{p, q}(P^n) + h^{p - 1, q - 1}(P^n) for p + q = n - 1 h^{p, q}(X) = h^{p, q}(P^n) for p + q > n - 1 In our case, X is a smooth hypersurface in P^3 of degree 4. The Hodge diamond of P^3 is given by: 1 0 0 1 0 1 0 0 0 1 Applying the Lefschetz hyperplane theorem, we get the Hodge diamond of X as: 1 0 0 1 0 1 0 0 0 1 So, the dimensions of the Hodge cohomology groups H^{p, q}(X) for p, q = 0, 1, 2, 3 are: H^{0, 0}(X) = 1 H^{0, 1}(X) = H^{1, 0}(X) = 0 H^{0, 2}(X) = H^{2, 0}(X) = H^{1, 1}(X) = 1 H^{0, 3}(X) = H^{3, 0}(X) = H^{1, 2}(X) = H^{2, 1}(X) = 0 H^{3, 1}(X) = H^{2, 2}(X) = H^{1, 3}(X) = 0 H^{2, 3}(X) = H^{3, 2}(X) = 0 H^{3, 3}(X) = 1

answer:The given equation is not a Calabi-Yau variety. Calabi-Yau varieties are complex manifolds with a vanishing first Chern class and a trivial canonical bundle, which implies that they have a Ricci-flat metric. They usually appear in the context of algebraic geometry and string theory. The equation you provided, y^2 = x^3 + 2x^2 + 1, is an elliptic curve, which is a one-dimensional complex torus (or a two-dimensional real torus). It is not a Calabi-Yau manifold, as Calabi-Yau manifolds have complex dimensions greater than or equal to two. However, if you are interested in the degree of the elliptic curve, it is given by the degree of the defining polynomial. In this case, the polynomial is y^2 - x^3 - 2x^2 - 1, which has degree 3 (due to the term x^3). So, the degree of this elliptic curve is 3.