ag.algebraic geometry - Degree of divisors and degrees of the corresponding maps to projective space

Here's how I think about it.

Let's assume we are in the case that $dimvarphi(X)=dim X$. Then $varphi : Xto varphi(X)$ is an generic finite map. Let $d$ be the degree of this map which is defined as the degree of field extension $[k(X):k(varphi(X))]$. The degree of $varphi(X)$ is given by $varphi(X)cdot H^{dim X}$ where $H$ is a general hyperplane of $mathbb{P}^n$. Pulling $H$ back to $X$, we get $D$. Then, by projection formula, $D^{dim X} = Xcdot D^{dim X}=dcdot(varphi(X)cdot H^{dim X})$. In the case that $X$ is a curve, $D^{dim X}$ is noting but the degree of $D$. So, the degree of $D$ equals that the degree of image times the degree of the map.

However, in higher dimension, $D^{dim X}$ may not be the degree of $D$. For example, $D$ is a irreducible degree 2 curve in $mathbb{P}^2$. The degree of $D$ is 2 which is not equal to $Dcdot D=4$ by Bézout's theorem.

Edit: I think in higher dimension, to define the degree of a divisor $D$, we need to choose a very ample divisor $A$ at first and then define the degree as the intersection number $Dcdot A^{dim D}$.