Let \(f: X \to Y\) be a finite morphism of curves of degree \(n\). We define a homomorphism \(f_*: \Div X \to \Div Y\) by \(f_*(\sum n_i P_i)=\sum n_i f(P_i)\) for any divisor \(D=\sum n_i P_i\) on \(X\).
For any locally free sheaf \(\mce\) on \(Y\), of rank \(r\), we define \(\det \mce=\wedge^r \mce \in \Pic Y\) (II, Ex. 6.11). In particular, for any invertible sheaf \(\mcm\) on \(X\), \(f_* \mcm\) is locally free of rank \(n\) on \(Y\), so we can consider \(\det f_* \mcm \in \Pic Y\). Show that for any divisor \(D\) on \(X\), \[ \det\left(f_* \mcl(D)\right) \cong \det f_*(\OO_X) \tensor \mcl(f_* D). \] Note in particular that \(\det(f_* \mcl(D)) \neq \mcl(f_* D)\) in general! Hint: First consider an effective divisor \(D\), apply \(f_*\) to the exact sequence \(0 \to \mcl(-D) \to \OO_X \to \OO_D \to 0\), and use (II, Ex. 6.11).
Conclude that \(f_* D\) depends only on the linear equivalence class of \(D\), so there is an induced homomorphism \(f_*: \Pic X \to \Pic Y\). Show that \(f_* f^*: \Pic Y \to \Pic Y\) is just multiplication by \(n\).
Use duality for a finite flat morphism (III, Ex. 6.10) and (III, Ex. 7.2) to show that \[\det f_* \Omega_X \cong \qty{ \det f_* \OO_X}^{-1} \tensor \Omega_Y^{\tensor n}.\]
Now assume that \(f\) is separable, so we have the ramification divisor \(R\). We define the branch divisor \(B\) to be the divisor \(f_* R\) on \(Y\). Show that \[\left(\det f_* \OO_X\right)^2 \cong \mcl(-B).\]
Solution
Write \[ E=f_*\OO_X. \] Since \(f\) is a finite morphism between nonsingular curves, it is flat; hence \(E\) is locally free of rank \(n\). More generally, \(f_*\mcm\) is locally free of rank \(n\) for every invertible sheaf \(\mcm\) on \(X\).
1. Let \(Q\in Y\). If\[ 0\longrightarrow E_0\longrightarrow E_1\longrightarrow k(Q) \longrightarrow0 \]is exact with \(E_0,E_1\) locally free of the same rank, then\[ \det E_1 \cong \det E_0\tensor\mcl(Q). \]
Away from \(Q\) the two bundles are equal. At \(Q\), let \[ A=\OO_{Y,Q} \] with uniformizer \(t\). After choosing bases, \(E_{0,Q}\subset E_{1,Q}\) are free \(A\)-lattices of the same rank and the quotient has length one. The elementary divisor theorem over the DVR \(A\) gives bases in which \[ E_{1,Q}=A^r, \qquad E_{0,Q}=tA\oplus A^{r-1}. \] Thus the determinant lattice of \(E_1\) is obtained from that of \(E_0\) by multiplying by \(t^{-1}\). This is exactly the local modification defining \(\mcl(Q)\), so \[ \det E_1\cong\det E_0\tensor\mcl(Q). \]
2. For every divisor \(D\) on \(X\) and every closed point \(P\in X\),\[ \det f_*\mcl(D+P) \cong \det f_*\mcl(D)\tensor\mcl(f(P)). \]
There is a short exact sequence \[ 0 \longrightarrow \mcl(D) \longrightarrow \mcl(D+P) \longrightarrow k(P) \longrightarrow0. \] Because \(f\) is finite, \(f_*\) is exact on quasicoherent sheaves. Since the ground field for the curves is algebraically closed, \[ f_*k(P)=k(f(P)) \] as a skyscraper sheaf. Applying \(f_*\) therefore gives \[ 0 \longrightarrow f_*\mcl(D) \longrightarrow f_*\mcl(D+P) \longrightarrow k(f(P)) \longrightarrow0. \] Step 1 gives the claimed determinant relation.
3. For every divisor \(D\) on \(X\),\[ \boxed{ \det\left(f_*\mcl(D)\right) \cong \det(f_*\OO_X)\tensor\mcl(f_*D). } \]
2. To subtract a point, apply step 2 to\(D-P\):\[ \det f_*\mcl(D-P) \cong \det f_*\mcl(D)\tensor\mcl(-f(P)). \]Iterating over all coefficients of\[ D=\sum_P n_P P \]gives\[ \det f_*\mcl(D) \cong \det E \tensor \mcl\left(\sum_Pn_Pf(P)\right), \]which is exactly the formula in part (a).
4. If \(D\sim D'\) on \(X\), then\[ f_*D\sim f_*D' \]on \(Y\). Hence \(f_*\) descends to a homomorphism\[ f_*:\Pic X\longrightarrow\Pic Y. \]
Linear equivalence \(D\sim D'\) is equivalent to \[ \mcl(D)\cong\mcl(D'). \] Applying \(f_*\) preserves this isomorphism, so their determinants are isomorphic. Step 3 gives \[ \det E\tensor\mcl(f_*D) \cong \det E\tensor\mcl(f_*D'). \] Cancelling \(\det E\) yields \[ \mcl(f_*D)\cong\mcl(f_*D'), \] which is the desired linear equivalence. Since the divisor pushforward is additive, the induced map on divisor classes is a group homomorphism.
5. The composite\[ f_*f^*:\Pic Y\longrightarrow\Pic Y \]is multiplication by \(n\).
It is enough to check a closed point \(Q\in Y\). The pullback divisor is \[ f^*Q = \sum_{P\mapsto Q}e_P P. \] Therefore \[ f_*f^*Q = \left(\sum_{P\mapsto Q}e_P\right)Q. \] For a finite morphism of degree \(n\) between curves over an algebraically closed field, the degree of every fibre counted with multiplicity is \(n\): \[ \sum_{P\mapsto Q}e_P=n. \] Thus \[ f_*f^*Q=nQ. \] By additivity this holds for every divisor and hence every divisor class. Equivalently, on line bundles the composite sends \(\mcm\) to \(\mcm^{\tensor n}\).
6. Finite duality gives an isomorphism\[ f_*\Omega_X \cong \mathcal Hom_Y(E,\Omega_Y) \cong E^\vee\tensor\Omega_Y. \]
By Exercise III.6.10 and Exercise III.7.2, finite duality identifies the pushforward of the dualizing sheaf of \(X\) with the dual of \(f_*\OO_X\) relative to the dualizing sheaf of \(Y\): \[ f_*\omega_X \cong \mathcal Hom_Y(f_*\OO_X,\omega_Y). \] On a nonsingular curve the dualizing sheaf is the sheaf of differentials, so \[ \omega_X\cong\Omega_X, \qquad \omega_Y\cong\Omega_Y. \] Because \(E\) is locally free, \[ \mathcal Hom_Y(E,\Omega_Y) \cong E^\vee\tensor\Omega_Y, \] which proves the displayed formula.
7. One has\[ \boxed{ \det f_*\Omega_X \cong (\det f_*\OO_X)^{-1}\tensor\Omega_Y^{\tensor n}. } \]
6. Since \(E\) has rank \(n\),\[ \det(E^\vee\tensor\Omega_Y) \cong \det(E^\vee)\tensor\Omega_Y^{\tensor n}. \]Also\[ \det(E^\vee)\cong(\det E)^{-1}. \]Substituting \(E=f_*\OO_X\) gives exactly the formula in part (c).
8. Assume now that \(f\) is separable, with ramification divisor \(R\) andbranch divisor\[ B=f_*R. \]Then\[ \Omega_X \cong f^*\Omega_Y\tensor\mcl(R). \]
For a finite separable morphism of nonsingular curves, the differential \[ f^*\Omega_Y\longrightarrow\Omega_X \] is a nonzero map of line bundles. Its divisor of zeros is the ramification divisor \(R\). A nonzero map of line bundles whose zero divisor is \(R\) identifies the target with the source tensored by \(\mcl(R)\), giving \[ \Omega_X \cong f^*\Omega_Y\tensor\mcl(R). \]
9. The determinant of \(E=f_*\OO_X\) satisfies\[ \boxed{ (\det E)^2\cong\mcl(-B). } \]
By step 8 and the projection formula, \[ f_*\Omega_X \cong \Omega_Y\tensor f_*\mcl(R). \] Taking determinants and using step 3 with \(D=R\) gives \[ \begin{aligned} \det f_*\Omega_X &\cong \Omega_Y^{\tensor n}\tensor\det f_*\mcl(R)\\ &\cong \Omega_Y^{\tensor n}\tensor\det E\tensor\mcl(f_*R)\\ &= \Omega_Y^{\tensor n}\tensor\det E\tensor\mcl(B). \end{aligned} \] On the other hand, step 7 gives \[ \det f_*\Omega_X \cong (\det E)^{-1}\tensor\Omega_Y^{\tensor n}. \] Cancel \(\Omega_Y^{\tensor n}\) and tensor by \(\det E\). The result is \[ (\det E)^2\tensor\mcl(B)\cong\OO_Y, \] or equivalently \[ (\det E)^2\cong\mcl(-B). \] This proves part (d).
10. Q.E.D.
Steps <1>1–<1>3 prove part (a), steps <1>4–<1>5 prove part (b), steps <1>6–<1>7 prove part (c), and steps <1>8–<1>9 prove part (d).