Let $L$ be a line bundle on a smooth curve $C$. If $L$ is rank $r+1$, define the induced map (as Arbarello, Cornalba, Griffiths, Harris):

$$\begin{aligned}\phi :& C \rightarrow \mathbb P|L|^*\\&p \mapsto \{s\in |L|:s(p) =0\}\end{aligned}$$

Let $B$ be the base locus of $L$, and let $C'$ denote the image of the above map. Why does it follow from this that $\deg(L)=\deg(C')\deg(\phi) + \deg(B)$?

ACGH states that this follows immediately from the local description of the above map, but I don't see how. Could someone please elaborate? Many thanks in advance.


1 Answer 1


Let us suppose $L$ is base-point-free first. Let $\phi_1:C\to C'$ be the restricted morphism. Now $\phi$ is a finite morphism of curves. By construction, $$ L=\phi^\ast(\mathscr O(1))=\phi_1^\ast \mathscr O(1)|_{C'}\,\Rightarrow\,\deg L=(\deg\phi_1)(\deg C')=(\deg\phi)(\deg C'). $$ The base locus of $\phi$ can be viewed as the effective divisor $B=\sum_{P\in C}n_P[P]$ where all sections in $H^0(C,L)$ vanish. The degree $\deg L$ is the number of zeros minus the number of poles of any rational section of $L$. But the rational map $\phi:C\dashrightarrow \mathbb P(|L|)^\vee$ is undefined exactly at the points $P$ in the support of $B$. So any such point has to contribute to $\deg L$ (necessarily by $n_P$).

Added. More on $\deg \phi_1^\ast \mathscr O(1)|_{C'}=(\deg\phi_1)(\deg C').$

For a finite morphism of curves $f:C\to C'$ and for any divisor $D\subset C'$, we have $f_\ast[f^\ast D]=(\deg f)[D]$, as cycles on $C'$. In particular, we have $\deg f_\ast[f^\ast D]=(\deg f)(\deg D)$. Let me be more precise: $$ \deg f^\ast[D]:=\int_Cf^\ast[D]=\int_{C'}f_\ast[f^\ast D]=\int_{C'}(\deg f)[D]=(\deg f)\int_{C'}[D]=:(\deg f)(\deg D). $$ For us, $D=\mathscr O(1)|_{C'}$, which has degree $\deg C'$ (I use, here, the definition of degree as the number of points in the intersection $C'\cap H$ with a general hyperplane: it seems more natural in this context, as $\mathscr O(1)$ corresponds to a hyperplane, and $\mathscr O(1)|_{C'}$ corresponds to intersecting $C'$ with a hyperplane).


Apart from the equality $f_\ast[f^\ast D]=(\deg f)[D]$, the only nontrivial issue in the displayed equations is the second equality. There, we use functoriality of degree. The degree of a cycle on a (proper) variety $Y$ can be defined (Fulton, Intersection Theory) as the operator $$\deg_Y(\cdot)=\int_Y\cdot=p_\ast(\cdot)\,\,\,\,\,\,\,\,\,\textrm{(these are synonyms!)}$$ Functoriality then says that if we have a morphism $f:X\to Y$ (with $X$ proper) then for a cycle $\alpha$ on $X$ we have $\deg_X(\alpha)=\deg_Y(f_\ast\alpha)$.

  • $\begingroup$ Dear Brenin, thank you for your answer. I understand the contribution of $\deg B$ to $\deg L$, but could you please elaborate on the equation $\deg L = (\deg \phi_1)(\deg C')$? The definitions that I know of degree of a line bundle is the degree of any of its sections. For the degree of a curve C, I know that it is $(\deg(P_C))!$ times the leading coefficient of $P_C$, where $P_C$ is the Hilbert polynomial of C. Finally, if $\phi(x) = (z_1(x),...,z_{r+1}(x))$, I understand $\deg(\phi) = \max (\deg(z_i))$. $\endgroup$
    – Rodrigo
    Commented Nov 19, 2013 at 16:28
  • $\begingroup$ Brenin, many, many thanks to your answer. But somehow I am still missing out on many things here. For example, I've never seen the degree be defined as an integral. For me, it was always just the sum of $n_p$. Can you provide any references? $\endgroup$
    – Rodrigo
    Commented Nov 20, 2013 at 2:13
  • $\begingroup$ @Rodrigo: you are welcome. Sorry if I did not choose the most neat way to explain the situation. But I added a reference. Btw, this coincided with the sum of the $n_p$ in good situations. $\endgroup$
    – Brenin
    Commented Nov 20, 2013 at 15:47

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