Degree of the pull-back of zero-cycles

Question

Let $f\colon X \to Y$ be a finite, flat, surjective morphism between smooth projective varieties over a field (not necessarily algebraically closed). Let $D$ be a $0$-cycle (meaning a zero dimensional cycle). Is $\deg(f^*D)=\deg(f)\cdot \deg(D)$? Any reference for this will be most welcome.

Answer 1

I think the following argument works, using the fact that $f$ is a projective morphism. Note that it suffices to prove the equality for $D=[p]$, where $p$ is a point of $Y$. Note also by flatness (of constant relative dimension $0$) that $f^*[p] = [f^{-1}(p)]$.

Since $f$ is flat, the Hilbert polynomial $h(X_p)$ of your fibres $X_p$ is constant as you vary $y\in Y$ . From Number of points in the fibre and the degree of field extension, we deduce that $h(X_p) = \mathrm{deg}(f)$ for an open subset of $Y$, and hence for all $y\in Y$. But $X_p$ is just $f^{-1}(p)$, and you can check that this is the same as saying that $\mathrm{deg} [f^{-1}(p)] = \mathrm{deg}(f)$.

Note that to use the proposition given in the link however I think you do need that the morphism is separable.

Here is another argument using more machinery from intersection theory (which generalises to much higher generality). We may write $D = D\cap [Y]$, where $\cap: A^p Y \otimes A_q X \rightarrow A_{q-p}X$ is defined as in Chapter 8 of Fulton's intersection theory. So we have $$\mathrm{deg} (f^*D) = \mathrm{deg} (f^*D \cap [X]) = \mathrm{deg} (f_* (f^*D \cap [X])) = \mathrm{deg} ( D \cap f_* [X] ) = \mathrm{deg} (D \cap \deg(f) [Y]) = \deg(f) \mathrm{deg} (D\cap [Y]) = \mathrm{deg}(f) \mathrm{deg}(D)$$

where the third equality is the projection formula.