Yesterday, we defined cycles, cycle classes, and the abelian groups in which they live.  Today, we’re going to fiddle with them a bit.  We’ve got a proper pushforward map, so today, we’ll start by figuring out when we have a pullback.

That question turns out to be less difficult than expected.  The key condition we’re really going to need is that we have fibers all the same dimension.  We’re going to assume that we actually have f:X\to Y a FLAT morphism, and that the fibers are all of dimension n.  This will give us pullbacks to open subschemes, to fiber bundles, to products with other schemes, and from a curve to a variety with a dominant map to the curve, so this will work out in a lot of the cases we care about.

So, very simple, we define f^*[V]=[f^{-1}(V)].  This gives us a map f^*:Z_k(Y)\to Z_{k+n}(X).  In fact, this definition will actually work for any subscheme, when we interpret f^{-1}(V) as the inverse image scheme of V under f.  That is, the fiber product of the scheme V with X over Y, both using the map f.  It’s a fact, though a bit technical to prove, that this map descends to A_k(Y)\to A_{k+n}(X).

Now, we’ve got that A_* is a covariant functor for proper morphisms and a contravariant functor for flat morphisms.  Now, we can use the pushforward and pullback operations to get an exact sequence: let Y be a closed subscheme of X, and U=X\setminus Y, with the inclusion maps i:Y\to X and j:U\to X.  Then we have an exact sequence for all k given by A_kY\stackrel{i_*}{\to}A_kX\stackrel{j^*}{\to}A_kU\to 0.  This is just the statement that cycle classes on U are just cycle classes on X which aren’t cycle classes on Y, which is something that is known to be true for divisors, that is, if X is a variety, Y an effective divisor, and U the complement, then the Picard group of U is just that of X after killing the class of Y.

Now, we look at affine bundles.  These are fiber bundles with fiber \mathbb{A}^n which trivialize in the Zariski topology.  Here, there’s a theorem: flat pullback is surjective for all k.  To prove it, we take U to be an open subset of X on which an affine bundle E is trivial, and Y the complement.  Applying the exact sequence, and then getting the maps via flat pullback, we can reduce to checking the problem on the inverse images of Y and U.  Induction on dimension takes care of Y, and on U, we have just U\times\mathbb{A}^n.  As the projections factor, we can even reduce to the case of U\times \mathbb{A}^1\to U, and this case is manageable.

Now, this implies that for \mathbb{A}^n, we have A_k(\mathbb{A}^n) zero, except for A_n(\mathbb{A}^n)=\mathbb{Z}, as it is an affine bundle over a point.  Now, note, we’re talking about affine bundles, not vector bundles.  The fibers are torsors rather than vector spaces.  When we have the additional structure of a vector space (and thus can projectivize and keep working) we’ll be able to prove that flat pullback is an isomorphism.

Here’s an example by Chow.  Let X has a cellular decomposition.  Say, like this one.  Then, we actually get that A_*X is finitely generated, and by the closures of the cells! So now, we know the Chow groups of the Grassmannian.

Finally, for our basic manipulations before we start trying to really do some intersections, let X,Y be schemes and X\times Y their product.  We have an exterior product on Chow groups given by [V]\times[W]=[V\times W].  This is a priori defined only on cycles, but, as always, preserves rational equivalence, and so descends to the Chow groups themselves.  Additionally, both pullback and pushforward distribute over this product, and the product is associative.  So we have a lot of nice properties for a product A_k(X)\otimes A_\ell(Y)\to A_{k+\ell}(X\times Y).  This should give us some hope there there is a ring structure nearby.  A quick guess would by that if \Delta:X\to X\times X, the diagonal, is flat (which it is) and of relative dimension zero, we should be able to get a graded ring structure by A_k(X)\otimes A_\ell(X)\stackrel{\times}{\to}A_{k+\ell}(X\times X)\stackrel{\Delta^*}{\to}A_{k+\ell}(X).