Scheme-theoretic preimages do not define a pullback on Chow groups

The geometric set-up. Choose homogeneous coordinates $[x:y:z]$ with $p=[0:0:1]$. The incidence subscheme $H=\{xv=yu\}\subset {\mathbb{P}}_k^2\times {\mathbb{P}}_k^1$, with coordinates $[u:v]$ on the second factor, is the blowup: on $z\mathrm{\ne }0$, its $u\mathrm{\ne }0$ and $v\mathrm{\ne }0$ charts are $\mathrm{Spec}k[x,v\mathrm{/}u]$ with $y=x(v\mathrm{/}u)$ and $\mathrm{Spec}k[y,u\mathrm{/}v]$ with $x=y(u\mathrm{/}v)$, the two Rees charts for $(x,y)$; away from $p$ the incidence projection is an isomorphism. These identifications glue to $H\cong Y$. The Segre embedding The Segre-Veronese map is a closed embedding therefore embeds $Y$ as a closed subscheme of projective space; the incidence embedding also proves that $\pi$ is projective. By [F2] the blowup $\pi :Y\to X$ at the rational point $p$ is a birational morphism of smooth projective surfaces, its exceptional curve $E={\pi }^{-1}(p)$ is isomorphic to ${\mathbb{P}}_k^1$, and $\pi$ restricts to an isomorphism ${\pi }^{-1}(X\setminus \{p\})\to X\setminus \{p\}$; in particular for the second rational point $q\mathrm{\ne }p$ the scheme-theoretic preimage ${\pi }^{-1}(q)=\{q^{\mathrm{\prime }}\}$ is a single reduced point.

The rationally equivalent cycles. In $A_0(X)\cong \mathbb{Z}$ the classes of $k$-rational points are all equal because the degree homomorphism sends each to $[\kappa (x):k]=1$ and is injective: $[p]=[q]$ in $A_0(X)$. The preimage cycles, however, lie in different Chow degrees: ${\pi }^{-1}(p)=E$ is a curve, so $[E]\in Z_1(Y)$, while ${\pi }^{-1}(q)=\{q^{\mathrm{\prime }}\}$ is a point, so $[q^{\mathrm{\prime }}]\in Z_0(Y)$.

The contradiction. If the preimage recipe descended to a homomorphism $\phi :A_{\ast }(X)\to A_{\ast }(Y)$, then $[p]=[q]$ would force $\phi [p]=\phi [q]$, that is $[E]=[q^{\mathrm{\prime }}]$ in $A_{\ast }(Y)$. Apply the proper pushforward ${\pi }_{\ast }$, which is well defined on rational equivalence by [F4]. Since $\pi (E)=\{p\}$ has dimension $0<1=\dim E$, the norm-degree formula gives ${\pi }_{\ast }[E]=0$; and since $\pi$ is an isomorphism over $q$, ${\pi }_{\ast }[q^{\mathrm{\prime }}]=[q]$. Together with $[E]=[q^{\mathrm{\prime }}]$ this gives $[q]=0$ in $A_0(X)$, contradicting that $[q]$ has degree $1$ by [F3]. By contrast the flat case of [F5] is well defined on Chow groups, so the failure is exactly the dimension jump of the non-flat morphism. Hence the preimage cycles of the rationally equivalent cycles $[p]$ and $[q]$ are not rationally equivalent, and the scheme-theoretic preimage recipe is not well defined on Chow groups.