$\det (f_{*} \mathcal{O}_D) \simeq \mathscr{L} (f_{*} D)$ (Hartshorne Exercise IV.2.6)

Question

I am stuck at the end of exercise IV.2.6 and I would appreciate a hint on how to conclude. We have $f:X\rightarrow Y$ a finite morphism of degree $n$ between curves and $D$ an effective divisor over X. I have to prove that $$\det(f_{*}\mathscr{L}(D))\simeq \det(f_{*}\mathcal{O}_X) \otimes \mathscr{L}(f_{*}D).$$ I managed to get to $$\det(f_{*}\mathscr{L}(D))\simeq \det(f_{*}\mathcal{O}_X) \otimes \det (f_{*} \mathcal{O}_D),$$ so I just need to prove that $\det (f_{*} \mathcal{O}_D) \simeq \mathscr{L} (f_{*} D)$, but I don't know how to proceed.

Answer 1

So first, if $D = \sum n_i p_i$, note that we can write $\mathcal{O}_D \cong \bigoplus_{p_i \in \operatorname{Supp} D} k(p_i)^{\oplus n_i}$ as a direct sum of skyscraper sheaves. This then implies that $f_*\mathcal{O}_D \cong \bigoplus_{p_i \in \operatorname{Supp} D} k(f(p_i))^{\oplus n_i} \cong \mathcal{O}_{f_*D}$.

But now, exercise II.6.11(b) says that the determinant of $\mathcal{O}_E$ is exactly $\mathscr{L}(E)$, for any effective divisor $E$ on a smooth curve. The desired result follows.

In case you haven't done this exercise, the determinant of a coherent sheaf $\mathscr{F}$ is defined to be $\operatorname{det} \mathscr{F} \cong \Lambda^{r_0}\mathscr{E}_0 \otimes (\Lambda^{r_1} \mathscr{E}_1)^{-1} \in \operatorname{Pic} X$ via a presentation of locally free sheaves. $$0 \to \mathscr{E}_1 \to \mathscr{E}_0 \to \mathscr{F} \to 0$$ In the case of $\mathcal{O}_D$, we have a presentation $$0 \to \mathscr{L}(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$$ which immediately gives us that $\operatorname{det}(\mathcal{O}_D) \cong \mathscr{L}(D)$.