3.4.4 Localization
Suppose \(U\) is a Zariski open subset of \(X\) with complement \(Y\), and \(k\in \mathbb {Z}\). Write \(i:Y\to X\) and \(j:U\to X\) for the inclusions. We have an obvious exact sequence on the level of algebraic cycles:
The point is that, as I’ve already mentioned, \(Z_k Y\) is a subgroup of \(Z_k X\). On the other hand, \(j\) is flat, we have \(j^*([V]) = [j^{-1} V]\) gives us the third morphism in 3.27. Moreover, it’s easy to see that \(j^*\) is onto, because the closure of any \(k\)-dimensional subvariety of \(U\) is a \(k\)-dimensional subvariety of \(X\).
Since \(i\) is proper and \(j\) is flat, 3.27 gives us a sequence
The sequence 3.28 is exact.
This is [ 3 , Proposition 1.8 ] .
The exactness of 3.27 makes it clear that \(j^*\) is onto. To show that 3.28 is exact, suppose \(j^*\alpha = 0\in A_k U\). We need to show that \(j^*=i_*\beta \) for some \(\beta \in Z_k Y\).
The hypothesis means that \(j^*\alpha = \sum _i \operatorname{\mathrm{div}}r_i\), where \(r_i\in K(W_i)\) for some \(k+1\)-dimensional subvarieties \(W_i\) of \(U\). But then let \(\bar W_i\) denote the closure of \(W_i\) in \(X\). Then \(K(\bar W_i) = K(W)\). So \(\operatorname{\mathrm{div}}r_i\) gives a cycle in \(Z_k X\), and we get that \(\alpha - \sum \operatorname{\mathrm{div}}r_i\) is supported on \(Y\). But that proves the theorem.