Let $\varphi\colon X_1\to X_2$ be dominant proper morphism of finite degree (in particular $\dim X_1=\dim X_2$) between varieties.

Let $D \subset X_2$ be a Cartier divisor.

Is it true that $$\varphi_*([\varphi^*(D)])=(\deg\varphi)[D]?$$ Here $[E]$ is the Weil divisor corresponding to $E$.

Remark: pullback on Cartier divisors is defined for any dominant morphism and pushforward on Weil divisors is defined for any proper morphism. So the formula make sense.

I do not assume varieties to be smooth nor the morphism to be flat. So for example $X_2$ can be singular at $D$.

In [Liu02, Proposition 9.2.11] it is proved for projective morphism between surfaces. In what generality is this statement true?

Q. Liu. Algebraic geometry and arithmetic curves

What do you think about this statement? Is it something standard? Or we need to assume some additional condition?

  • 1
    $\begingroup$ Yes, this is indeed true. If $X_1$ and $X_2$ are normal then this is clear since one can then easily reduce to the smooth case by removing a subvariety of codimension $2$ (which does not change Weil divisors). So the main case is when $X_1$ is the normalisation of $X_2$. But in this case the formula just amounts to the definition of the Weil divisor associated to a Cartier divisor (see Example 1.2.3 of Fulton's "Intersection Theory"). $\endgroup$
    – naf
    Oct 16, 2023 at 2:54
  • $\begingroup$ Do you know a reference? I want to cite it. $\endgroup$ Oct 16, 2023 at 7:34
  • $\begingroup$ You can view the statement as the special case of the projection formula, Proposition 2.3 (c) in Fulton's book, with $f$ there being your $\phi$ and $\alpha$ there being the class $[X_1]$, so $\phi_*([X_1]) = \deg(\phi)[X_2]$. $\endgroup$
    – naf
    Oct 16, 2023 at 15:25
  • $\begingroup$ Yes, thank you. This is exactly what I want! $\endgroup$ Oct 17, 2023 at 19:22


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