Let $k$ be an algebraically closed field and let $\mathcal{C}$ be proper, Cohen-Macaulay, purely $1$-dimensional Deligne-Mumford stack over $k$. From looking at section 4.3 on page 135 of the paper "Les schemas de modules des courbes elliptiques" by P. Deligne and M. Rapoport, I understand that there is a notion of degree for line bundles $\mathscr{L}$ on $\mathcal{C}$. My question is about some of the compatibility claims left to the reader in that reference and has several subquestions listed below.

The degree of $\mathscr{L}$ is defined starting from a "rational section" $f$ of $\mathscr{L}$. My first question is:

- What precisely is $f$? How is it defined?

The only reasonable candidate in my mind is a rational section of the pushforward to $\mathscr{L}$ to the coarse space of $\mathcal{C}$, but does this really work? For instance, why can't it happen that this pushforward is zero?

Once we have $f$, we look at strict Henselizations of $\mathcal{C}$ at closed points $x$ and, letting $R_x$ be the coordinate algebra of such a Henselization, we look at the "order" $\mathrm{deg}_x(f)$ with respect to $R_x$ of the pullback of $f$ to the total ring of fractions of $R_x$. In the text it is implicitly claimed that either the pullback of $f$ or its inverse is actually in $R_x$; I cannot see why this is so, but the order makes good sense nevertheless by expressing the pullback as a fraction (presumably at this stage one sees that the Cohen-Macaulayness condition cannot actually be dropped?). One finally defines $$ \mathrm{deg}(\mathscr{L}) = \sum_x \frac{1}{|\mathrm{Aut}(x)|} \mathrm{deg}_x(f).$$

My other subquestions are:

- Why is $\mathrm{\deg}(\mathscr{L})$ independent of $f$?
- Why is $\mathrm{\deg}(\mathscr{L})$ invariant under base change to an algebraically closed overfield (I bet this one is easy, but I cannot quite see it)?
- Why is $\mathrm{\deg}(\mathscr{L})$ "invariant under specialization", as claimed in $\gamma)$ of the reference? In other words, if $S$ is a spectrum of a DVR and $\widetilde{\mathcal{C}}$ and $\widetilde{\mathscr{L}}$ are over $S$, why is $\mathrm{deg}(\mathscr{\widetilde{L}}_\eta) = \mathrm{deg}(\mathscr{\widetilde{L}}_s)$?

I apologize for so many multiple parts, but comments on any one of them would help.

rationalPicard groups, with inverse $\pi_*$. So $\deg\pi_*(\mathcal L)$ is a well defined rational number. This works more generally for Chern classes of vector bundles. $\endgroup$ – Niels Feb 9 '16 at 10:30