# computing the degree of a line bundle

Let $C$ be a curve over a field $k$, $T$ a $k$-scheme.

Let $D$ be an effective divisor of degree $d$ on $C\times T$ parametrized by $T$. This means that $D$ is a closed subscheme of $C\times T$ s.t. for $i:D\rightarrow C\times T$ the inclusion, $\pi: C\times T\rightarrow T$ the projection, we have a $p: D\rightarrow T$ with $p=\pi\circ i$ and $p$ is finite and flat of degree $d$ (is it true that $p$ with $p=\pi\circ i$ is finite and flat of degree $d$ iff the vectorbundle $p_*(\mathcal{O}_D)$ has rank $d$?)

Let $s$ be a section of $p$. Then $s$ defines a closed subscheme of $D\subseteq C\times T$. The inclusion of the ideal sheaves $\mathcal{I}_D\hookrightarrow\mathcal{I}_{s(T)}$ gives rise to a line bundle $\mathcal{I}_D \otimes \mathcal{I}_{s(T)}^{-1}\hookrightarrow \mathcal{O}_{C\times T}$.

My question: Which degree does the effective divisor corresponding to the ideal sheaf $\mathcal{I}_D \otimes \mathcal{I}_{s(T)}^{-1}$ on $C\times T$ have? The solution should be $d-1$.

If I'm right, then I should show that the rank of $p_*(\mathcal{I}_D \otimes \mathcal{I}_{s(T)}^{-1})$ is $d-1$.

Thank you very much for all help!!!

• How are you defining the degree of a divisor on $C\times T$? Usually only '$0$-cycles' have a notion of degree. – Alex Youcis Mar 7 '15 at 21:57
• I'm very sorry, I improved my question. – Omega Mar 8 '15 at 13:49