By abuse of notation, we denote the canonical class again by. If is not smooth, can be neither injective nor surjective. Hence taking the vanishing locus of gives a well-defined Chern class map If is smooth, then is indeed an isomorphism since every codimension one cycle can be represented by a Cartier divisor. The vanishing locus and of two rational sections of are rationally equivalent since they can be connected by the family. Any line bundle on has a rational section. Let be a quasi-projective variety variety. On the other hand, this intersection can be compute directly (without push-pull formula), it is the number of ways of grouping six points into three pairs of points, which is. Let be general points, then Be careful that one needs to verify that this intersection is transverse since are no longer general hyperplanes and we cannot invoke Bertini's theorem directly.

For any point, we have a hyperplane of cubics passing through. The formula for then follows from the fact that One subtle point is that we need to actually write down local equations of to check that it is reducible, i.e., multiplicity-free. Let be the closure of the triples Then the fiber and is isomorphic to (as the graph of an automorphism of ). The second method is less general: specialization. A tiny proportion of varieties for which we can compute the Chow rings actually has this secret property). (The secret property we are using is that the cohomology ring and Chow ring agree for products of projective spaces. Hence This is reminiscence of the cohomology class of the diagonal in topology. This is the same as the intersection number, which is equal to 1. To determine, we take the intersection of with and observe that, by the shape of the Chow ring of, This is equal to, where and are general linear subspaces such that the intersection is transverse. The first method is undetermined coefficients, which applies many other settings. We observe that has an affine stratification The key to proving this is that every subvariety is rationally equivalent to a multiple of a linear subspace. Namely, if is of codimension, degree, then. Nevertheless this is not the end of the world: a lot of calculation are performed on spaces whose Chow rings are known (e.g., projective spaces product of projective spaces Grassmannians) we also know many subrings of Chow rings (of certain special degrees). In contrast to homology/cohomology, the Chow groups are not computed for 99 percent of algebraic varieties: even for simple cases like surfaces in of degree (for, the Chow ring is : every two points are connected by rational curves we don't know if this is true or not for ).

On the other hand, the cup product on cohomology doesn't need such restrictive assumption and recovers the intersection pairing on homology via Poincare duality for oriented smooth manifolds. It also follows the push-pull formula: for, , then Since pullback preserves codimension but not dimension, we will also use the grading by the codimension of cycles (only when is smooth) An analogy: the intersection pairing on homology is only well-defined on nice topological spaces ( = oriented smooth manifolds) since one needs a good notion of index (which can be negative) for components in the intersection. Then the moving lemma implies that there exists a well-defined pullback map, given by when and is generically reduced. For us, it does no real harm to take the moving lemma for granted and use it as a tool to do algebraic geometry. Fulton wrote an influential book in 1984 and figured out a way to prove the key fact without proving the moving lemma (which implies the moving lemma) and put the theory on a solid basis.

Everybody believed that the second part follows easily from the first part, which is however not the case. Severi had a wonderful idea of the proof for the first part. The reason for this key fact is the moving lemma, which says that, firstly, for any classes, there exist cycles representing them and intersecting generically transversely secondly, is independent of this choice. In other words, the intersection behaves well under rational equivalence: Bezout's theorem is simply a tiny instance of this phenomenon. The key fact is that for smooth and projective, one can define a product on such that for any intersecting generically transversely (i.e., the intersection is transverse at a generic point of every component of ) such that.

Subspace definition math