Let \(X\) and \(X'\) be elliptic curves over \(k\), with base points \(P_0, P_0'\).
If \(f: X \to X'\) is any morphism, use (4.11) to show that \(f^*: \Pic X' \to \Pic X\) induces a homomorphism \(\hat{f}: (X', P_0') \to (X, P_0)\). We call this the dual of \(f\).
If \(f: X \to X'\) and \(g: X' \to X''\) are two morphisms, then \(\widehat{g \circ f} = \hat{f} \circ \hat{g}\).
Assume \(f(P_0)=P_0'\), and let \(n=\deg f\). Show that if \(Q \in X\) is any point, and \(f(Q)=Q'\), then \(\hat{f}(Q')=n_X(Q)\). (Do the separable and purely inseparable cases separately, then combine.) Conclude that \(f \circ \hat{f}=n_{X'}\) and \(\hat{f} \circ f=n_X\).
* If \(f, g: X \to X'\) are two morphisms preserving the base points \(P_0, P_0'\), then \(\widehat{f+g} = \hat{f}+\hat{g}\).
Hints: It is enough to show for any \(\mcl \in \Pic X'\), that \((f+g)^* \mcl \cong f^* \mcl \tensor g^* \mcl\). For any \(f\), let \(\Gamma_f: X \to X \times X'\) be the graph morphism. Then it is enough to show (for \(\mcl'=p_2^* \mcl\)) that \[ \Gamma_{f+g}^*(\mcl')=\Gamma_f^* \mcl' \tensor \Gamma_g^* \mcl' . \] Let \(\sigma: X \to X \times X'\) be the section \(x \mapsto (x, P_0')\). Define a subgroup \(\Pic_\sigma\) of \(\Pic(X\times X')\) consisting of those \(\mcl\) which have degree 0 along each fibre of \(p_1\) and satisfy \(\sigma^* \mcl = 0 \in \Pic(X)\). Note that this subgroup is isomorphic to the group \(\Pic^{\circ}(X'/X)\) used in the definition of the Jacobian variety. Hence there is a 1-1 correspondence between morphisms \(f: X \to X'\) and elements \(\mcl_f \in \Pic_\sigma\) (this defines \(\mcl_f\)). Now compute explicitly to show that \(\Gamma_g^*(\mcl_f)=\Gamma_f^*(\mcl_g)\) for any \(f, g\). Use the fact that \(\mcl_{f+g}=\mcl_f \tensor \mcl_g\), and the fact that for any \(\mcl\) on \(X'\), \(p_2^* \mcl \in \Pic_\sigma^{\circ}\) to prove the result.
Using (d), show that for any \(n \in \ZZ\), \(\hat{n}_X=n_X\). Conclude that \(\deg n_X=n^2\).
Show for any \(f\) that \(\deg \hat{f}=\deg f\).
Solution
For an elliptic curve \((E,O)\) write \[ \alpha_E:E\longrightarrow\Pic^0(E), \qquad P\longmapsto\OO_E(P-O). \] The preceding results identify \(\alpha_E\) as an isomorphism of group varieties.
1. For every morphism\[ f:X\longrightarrow X' \]the pullback on line bundles induces a homomorphism\[ \boxed{ \hat f = \alpha_X^{-1}\circ f^*\circ\alpha_{X'} : (X',P_0')\longrightarrow(X,P_0). } \]
If \(f\) is nonconstant of degree \(m\), then \[ \deg f^*\mcl=m\deg\mcl, \] so pullback carries \(\Pic^0(X')\) into \(\Pic^0(X)\). If \(f\) is constant, the pullback of every degree-zero line bundle is trivial, so the same conclusion holds.
The relative Picard construction used in (4.11) shows that pullback is not merely a homomorphism on \(k\)-points: it is a morphism of the represented Picard varieties. Moreover \[ f^*(\mcl\tensor\mcn) \cong f^*\mcl\tensor f^*\mcn, \] so it is a group homomorphism. Conjugating by the group-variety isomorphisms \(\alpha_{X'}\) and \(\alpha_X\) gives the asserted homomorphism \(\hat f\). Since \[ f^*\OO_{X'}\cong\OO_X, \] it sends \(P_0'\) to \(P_0\).
2. If\[ X\xrightarrow{f}X'\xrightarrow{g}X'' \]are morphisms, then\[ \boxed{ \widehat{g\circ f} = \hat f\circ\hat g. } \]
Pullback of line bundles is contravariantly functorial: \[ (g\circ f)^*=f^*\circ g^*. \] Therefore \[ \begin{aligned} \widehat{g\circ f} &= \alpha_X^{-1}(g\circ f)^*\alpha_{X''}\\ &= \alpha_X^{-1}f^*g^*\alpha_{X''}\\ &= (\alpha_X^{-1}f^*\alpha_{X'}) (\alpha_{X'}^{-1}g^*\alpha_{X''})\\ &= \hat f\circ\hat g. \end{aligned} \] Here the middle juxtaposition denotes composition of the displayed morphisms.
3. Assume now that\[ f(P_0)=P_0', \qquad n=\deg f>0. \]If \(f\) is separable and\[ f(Q)=Q', \]then\[ \boxed{\hat f(Q')=[n]Q.} \]
Because \(f\) carries origin to origin, it is a homomorphism of elliptic curves. A separable morphism between genus-one curves is unramified by Riemann–Hurwitz. Hence \[ K=\ker f \] has exactly \(n\) points and \[ f^{-1}(Q')=Q+K \] with every point occurring with multiplicity one. Thus \[ \begin{aligned} f^*\OO_{X'}(Q'-P_0') &\cong \OO_X\left( \sum_{T\in K}(Q+T)-\sum_{T\in K}T \right). \end{aligned} \] Under the identification \[ \Pic^0(X)\cong X, \] the class of a degree-zero divisor \[ \sum_i P_i-\sum_i R_i \] corresponds to \[ \sum_iP_i-\sum_iR_i \] in the group law. The displayed divisor therefore corresponds to \[ \sum_{T\in K}(Q+T)-\sum_{T\in K}T = nQ. \] By the definition in step 1 this point is \(\hat f(Q')\).
4. Under the same hypotheses, if \(f\) is purely inseparable, then again\[ \boxed{\hat f(Q')=[n]Q.} \]
A finite purely inseparable morphism of degree \(n\) between smooth curves is radicial. Over the algebraically closed field, each fibre therefore has a single point, with scheme-theoretic multiplicity \(n\). Hence \[ f^*(Q')=nQ, \qquad f^*(P_0')=nP_0. \] Consequently \[ f^*\OO_{X'}(Q'-P_0') \cong \OO_X\bigl(nQ-nP_0\bigr), \] whose class corresponds to \([n]Q\).
5. For an arbitrary nonconstant \(f\) preserving the origins, one still has\[ \boxed{ \hat f(f(Q))=[n]Q } \]for every \(Q\in X\).
Factor the finite extension of function fields into its inseparable and separable parts. On smooth projective models this gives \[ X\xrightarrow{f_i}Y\xrightarrow{f_s}X', \] where \(f_i\) is purely inseparable of degree \(i\), \(f_s\) is separable of degree \(s\), and \[ n=is. \] The intermediate curve inherits the image of \(P_0\) as origin, and both maps preserve origins.
Put \[ R=f_i(Q). \] Steps <1>3–<1>4 and contravariance from step 2 give \[ \begin{aligned} \hat f(f(Q)) &= \widehat{f_s\circ f_i}(f_s(R))\\ &= \hat f_i\bigl(\hat f_s(f_s(R))\bigr)\\ &= \hat f_i([s]R)\\ &= [s]\hat f_i(R)\\ &= [s][i]Q\\ &= [n]Q. \end{aligned} \]
6. Consequently\[ \boxed{ \hat f\circ f=[n]_X, \qquad f\circ\hat f=[n]_{X'}. } \]
The first equality is exactly step 5.
Since a nonconstant morphism of projective curves is surjective, every \(Q'\in X'\) is \(f(Q)\) for some \(Q\in X\). Then \[ \begin{aligned} f(\hat f(Q')) &= f([n]Q)\\ &= [n]f(Q)\\ &= [n]Q', \end{aligned} \] because \(f\) is a group homomorphism. Hence \[ f\circ\hat f=[n]_{X'}. \] This proves part (c).
7. Let\[ p_1:X\times X'\longrightarrow X, \qquad p_2:X\times X'\longrightarrow X' \]be the projections and let\[ \sigma:X\longrightarrow X\times X', \qquad x\longmapsto(x,P_0'). \]For a morphism \(h:X\to X'\) define\[ \mcm_h = \OO_{X\times X'}\bigl(\Gamma_h-X\times\{P_0'\}\bigr) \tensor p_1^*h^*\OO_{X'}(P_0')^{-1}. \]Then \(\mcm_h\) represents \(h\) in the normalized relative Picard group\(\Pic_\sigma\) occurring in the hint.
On the fibre \(\{x\}\times X'\) the graph \(\Gamma_h\) cuts out the point \(h(x)\), while \(X\times\{P_0'\}\) cuts out \(P_0'\). Therefore \[ \mcm_h|_{\{x\}\times X'} \cong \OO_{X'}(h(x)-P_0'). \] The last factor is pulled back from \(X\) and hence does not change this fibre class.
It remains to check the normalization along \(\sigma\). The graph is the inverse image of the diagonal under \[ (h\circ p_1,p_2):X\times X'\longrightarrow X'\times X'. \] Restricting the diagonal bundle to \(X'\times\{P_0'\}\) gives \(\OO_{X'}(P_0')\). Hence \[ \sigma^*\OO(\Gamma_h) \cong h^*\OO_{X'}(P_0'). \] The restriction of \(\OO(X\times\{P_0'\})\) along the constant section is trivial as a line bundle on \(X\). Consequently \[ \sigma^*\mcm_h\cong\OO_X. \] Thus \(\mcm_h\) has degree zero on every \(p_1\)-fibre and is normalized along \(\sigma\). Its fibre class is exactly \[ x\longmapsto\OO_{X'}(h(x)-P_0'), \] so representability of \(\Pic^0(X'/X)\) identifies it with the morphism \(h\).
8. For any two morphisms \(f,g:X\to X'\) one has the symmetric identity\[ \boxed{ \Gamma_g^*\mcm_f \cong \Gamma_f^*\mcm_g. } \]
Pulling the graph divisor \(\Gamma_f\) back along \(\Gamma_g\) gives the zero divisor of the difference morphism \[ g-f:X\longrightarrow X'. \] Therefore \[ \Gamma_g^*\OO(\Gamma_f) \cong (g-f)^*\OO_{X'}(P_0'). \] Also \[ \Gamma_g^*\OO(X\times\{P_0'\}) \cong g^*\OO_{X'}(P_0'), \] and \[ \Gamma_g^*p_1^*f^*\OO_{X'}(P_0') \cong f^*\OO_{X'}(P_0'). \] Thus \[ \Gamma_g^*\mcm_f \cong (g-f)^*\OO(P_0') \tensor g^*\OO(P_0')^{-1} \tensor f^*\OO(P_0')^{-1}. \] Since inversion fixes \(P_0'\), \[ [-1]^*\OO_{X'}(P_0')\cong\OO_{X'}(P_0'), \] so \[ (g-f)^*\OO(P_0') \cong (f-g)^*\OO(P_0'). \] The resulting expression is symmetric in \(f\) and \(g\), proving the claim.
9. If \(f,g:X\to X'\) preserve the base points, then\[ \boxed{ \widehat{f+g}=\hat f+\hat g. } \]
The group law on the relative Jacobian is tensor product, so step 7 gives \[ \mcm_{f+g} \cong \mcm_f\tensor\mcm_g. \]
Let \(\mcl\in\Pic^0(X')\). There is a unique point \(A\in X'\) such that \[ \mcl\cong\OO_{X'}(A-P_0'). \] Let \[ h:X\longrightarrow X' \] be the constant morphism with value \(A\). By step 7, \[ \mcm_h\cong p_2^*\mcl; \] the possible normalization factor pulled back from \(X\) is trivial because \(h\) is constant.
Using step 8 three times, \[ \begin{aligned} (f+g)^*\mcl &= \Gamma_{f+g}^*\mcm_h\\ &\cong \Gamma_h^*\mcm_{f+g}\\ &\cong \Gamma_h^*\mcm_f \tensor \Gamma_h^*\mcm_g\\ &\cong \Gamma_f^*\mcm_h \tensor \Gamma_g^*\mcm_h\\ &= f^*\mcl\tensor g^*\mcl. \end{aligned} \] Under \[ \alpha_X:X\overset\sim\longrightarrow\Pic^0(X), \] tensor product is addition. Applying the definition of the dual from step 1 to the last identity yields \[ \widehat{f+g}(A) = \hat f(A)+\hat g(A). \] This holds for every \(A\in X'\), so the two morphisms are equal. This proves the starred part (d).
10. For every \(r\in\ZZ\),\[ \boxed{\widehat{[r]_X}=[r]_X.} \]
The identity morphism pulls every line bundle back to itself, so step 1 gives \[ \widehat{[1]_X}=[1]_X. \] By step 9, dualization is additive on endomorphisms. Hence it sends zero to zero, sends negatives to negatives, and sends the sum of \(r\) copies of the identity to the sum of \(r\) copies of its dual. Therefore \[ \widehat{[r]_X}=[r]_X \] for every integer \(r\).
11. For every \(r\in\ZZ\),\[ \boxed{\deg [r]_X=r^2.} \]
For \(r=0\) this is the convention that a constant morphism has degree \(0\). Let \(r\ne0\) and put \[ d=\deg[r]_X. \] Applying step 6 to \([r]_X\) and using step 10 gives \[ [d]_X = \widehat{[r]_X}\circ[r]_X = [r^2]_X. \] The standard embedding \(\ZZ\to\Endo(X)\), \(m\mapsto[m]_X\), is injective, so \(d=r^2\). This proves part (e).
12. For every morphism \(f:X\to X'\),\[ \boxed{\deg\hat f=\deg f.} \]
If \(f\) is constant, step 1 shows that \(\hat f\) is the zero morphism, so both degrees are \(0\). Suppose that \(f\) is nonconstant and put \[ n=\deg f>0. \] Step 6 gives \[ f\circ\hat f=[n]_{X'}. \] Degrees of finite morphisms of curves multiply under composition. Hence step 11 gives \[ n^2 = \deg[n]_{X'} = \deg(f\circ\hat f) = n\deg\hat f. \] Since \(n>0\), we obtain \[ \deg\hat f=n=\deg f. \] This proves part (f).
13. Q.E.D.
Steps <1>1–<1>2 prove parts (a)–(b), steps <1>3–<1>6 prove part (c), steps <1>7–<1>9 prove part (d), steps <1>10–<1>11 prove part (e), and step 12 proves part (f).