# Pull and push formula for degree for non-flat morphism

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

• 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").
• 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 $\phi_*([X_1]) = \deg(\phi)[X_2]$.