Let $f:X\to Y$ be a finite dominant morphism of nonsingular varieties over some algebraically closed field. Let $P\in X$ be a point of codimension one, then $f(P)$ is also of codimension one. We can define the ramification index $e\_P$ of $P$ in the same way one would do it if $X$ and $Y$ were curves, my main reference here is Hartshorne IV.2. My question is somewhat about the correct generalization of Proposition II.6.9 in Hartshorne.


From Liu's Book (Algebraic Geometry and Arithmetic Curves; in particular Thm. 7.2.18 and Exc. 7.2.3) I conclude that for any $Q\in Y$ of codimension one, we should obtain the equality 


$\deg(f)=\displaystyle\sum\_{P\in f^{-1}(Q)} e\_P\cdot [k(P):k(Q)]$.


while


$\deg(f^\ast Q) = \displaystyle\sum\_{P\in f^{-1}(Q)} e\_P$


Now, I really want $\deg(f^\ast Q)$ to be constant for all $Q$ of codimension one - is this always the case? If not, can you name additional assumptions to make this work? Clearly, the condition $[k(P):k(Q)]=1$ for all $f(P)=Q$ comes to mind - in other words, the restriction of $f$ to any prime divisor of $X$ is an isomorphism. Does this hold under certain assumptions about $f$?



When is the degree of the pull-back of a Weil divisor a constant multiple of its degree?
In the situation you consider the degree $\mathrm{deg}(f)$ is equal to the degree $[K(X):K(Y)]$ of the extension of the function fields of $X$ and $Y$. The precise requirements are that $X$ and $Y$ are normal varieties and that $f$ is finite. The base field is not required to be algebraically closed.


For varieties of dimension $\geq 2$ there is no degree function for Weil divisors comparable to the one in the case of curves. Your definition(?) of $\mathrm{deg}(f^\ast Q)$ looks at least strange to me. How would you define the degree of an arbitrary Weil (prime) divisor of $X$? There should then be no reference to $f$ ...


