3.4.5 Pullback for smooth schemes and the projection formula
Here I mainly want to list some facts because it’s hard to say much without taking a lot of time. Firstly, if \(X\) is smooth and equidimension of dimension \(n\), then we have a commutative ring structure on \(A^* X\). The unit is \([X]\). If \(f:X\to Y\) is a morphism of smooth varieties, then we get a ring homomorphism \(f^{!}:A^* Y\to A^* X\). This ring homomorphism agrees with flat pullback in the case that \(f\) is flat.
Suppose \(f:X\to Y\) is a proper morphism of smooth varieties. Then proper pushforward \(f_*:A^* X\to A^*Y\) is an \(A^*Y\)-module morphism, where \(A^*X\) get a module structure via \(f^!:A^*Y\to A^*X\). Explicitly, we have
for \(\alpha \in A^* X\) and \(\beta \in A^* Y\).
Suppose \(f:X\to Y\) is a morphism of smooth, projective varieties and \(D\) is a divisor in \(Y\). If \(D\) is numerically equivalent to \(0\), then so is \(f^!D\).
Suppose \(C\) is a curve in \(X\), and write \(a_Y:Y\to \operatorname{\mathrm{Spec}}k\) (resp. \(a_X:X\to \operatorname{\mathrm{Spec}}k\)) for the structure map. Then \(C\cdot f^! D = a_{X*}([C] f^!D) = a_{Y*}f_*([C] f^! D) = f_*[C]\cdot D = 0\), since \(D\) is numerically equivalent to \(0\).
It follows that the map \(f^!:A^1 Y\to A^1 X\) factors to give a map \(f^!:\operatorname{\mathrm{NS}}Y\to \operatorname{\mathrm{NS}}X\).