# Degree of a Cartier Divisor under pullback

This is question 7.2.3 in Liu's book Algebraic Geometry and Arithmetic Curves and I have been trying with this for some time now.

Let $f:X \rightarrow Y$ be a morphism of Noetherian schemes, and suppose that X and Y are integral and that f is finite surjective. We will let $Div(Y)$ resp. $Div(X)$ stand for Cartier Divisors on Y resp. X.

Associated to a Cartier divisor $D \in Div(Y)$ we have a notion of multiplicity at a point $y \in Y$, defined as $$\text{mult}_y D = \text{length } \mathcal{O}_{Y,y}/D_y \mathcal{O}_{Y,y}.$$

I am trying to show that for a Cartier Divisor $D \in Div(Y)$, we have that $$\text{mult}_x f^\ast(D) = \text{length } \mathcal{O}_{X,x}/(\mathcal{m}_y\mathcal{O}_{X,x}) \cdot \text{mult}_y(D)$$ where $m_y$ is the maximal ideal of $\mathcal{O}_{Y,y}$.

Now, it is quite easy to do this if f is a flat morphism, since we can simply use 50.13 of this document http://stacks.math.columbia.edu/download/algebra.pdf (Stacks, Algebra). But I don't see how to do it in the general case, and I have been trying for some time, so any hint would be nice.

Thank you!

• Dear Dedalus, I think you need to assume that $Y$ is of dimension one, otherwise the multiplicity will be infinite if $y \in D$ is a non-generic point. (And, without fixing a projective embedding, degree is only meaningfully defined for divisors in curves.) Regards, – Matt E May 18 '13 at 11:00
• Dear Matt E, thank you for your reply. I should have added that Liu considers multiplicity only for points of codimension 1. – Dedalus May 18 '13 at 11:04
• (I am still very much of a beginner in scheme theory so please excuse my ignorance.) – Dedalus May 18 '13 at 11:05
• Dear Dedalus, Thanks for the reply. One way to think about it is that degree is a concept for a projective variety. The only time it intrinsic is for points (when it just counts the number of points). In the setting of a codim'n one point on divisors in a higher dimensional varieties, the multiplicity counts the "number of times" the particular irred. component corresponding to that point appears in the divisor. Regards, – Matt E May 18 '13 at 20:38

I just e-mailed Qing Liu about this exercise and got a nice reply (as always), and I will share it in case other people try this exercise and get stuck. Please excuse that I refer to numbers in the book

"7.2.3 a) is incorrect for the case f finite surjective, see EGA IV 5.6.11 . Maybe it can be corrected by assuming further that Y is normal.

In b) one should keep the hypothesis that Y is normal. "

I think I know how to do this in this case and will post a solution then, by editing this answer.

Update:

OK - here is how I think one can do it in the case Y is normal. Let us note that the property is local on Y, so that we can assume Y to be a normal integral, local noetherian ring of dimension 1. As such, Y is a Dedekind scheme. Now, we know that for a non-constant morphism $f: X \rightarrow Y$ with X integral and Y Dedekind , f is flat (see Liu, Corollary 4.3.10) . Hence, it reduces to the flat case and this is an easy exercise in commutative algebra.

And I forgot to add, if I am wrong, please tell me!