Let \(X\) be an integral projective scheme of dimension 1 over \(k\), which is locally complete intersection, and has \(p_a=1\). Fix a point \(P_0 \in X_{\reg}\). Imitate (1.3.7) to show that the map \(P \mapsto \mcl(P-P_0)\) gives a one-to-one correspondence between the points of \(X_{\reg}\) and the elements of the group \(\Pic X\).
This generalizes (II, 6.11.4) and (II, Ex. 6.7).
Erratum: the target group is \(\Pic^0 X\), not \(\Pic X\). The sheaf \(\mcl(P-P_0)\) has degree \(0\), so the map \(P \mapsto \mcl(P-P_0)\) lands in the subgroup \(\Pic^0 X\) of invertible sheaves of degree \(0\), and the one-to-one correspondence is between the points of \(X_{\reg}\) and \(\Pic^0 X\) [@Har10a, Exercise IV.1.10].
Solution
We prove the corrected statement with target \(\Pic^0 X\).
1. The canonical divisor \(K\) from Exercise IV.1.9 satisfies\[ \deg K=0, \qquad \ell(K)=1. \]
Since \(p_a(X)=1\), the singular-curve Riemann–Roch formula Exercise IV.1.9 gives, for \(D=0\), \[ \ell(0)-\ell(K) = 1-p_a(X) = 0. \] Because \(\ell(0)=h^0(X,\mco_X)=1\), this gives \(\ell(K)=1\).
Apply the same formula with \(D=K\): \[ \ell(K)-\ell(0) = \deg K+1-p_a(X) = \deg K. \] The left-hand side is zero, so \(\deg K=0\).
2. An invertible sheaf of negative degree on \(X\) has no nonzero globalsection.
Let \(\mcl\) be invertible and suppose \[ 0\ne s\in H^0(X,\mcl). \] Since \(X\) is integral, multiplication by \(s\) gives an injection \[ 0 \longrightarrow \mco_X \xrightarrow{s} \mcl \longrightarrow \mct \longrightarrow 0, \] where \(\mct\) is supported in dimension zero. Therefore \[ \deg\mcl = \chi(\mcl)-\chi(\mco_X) = \chi(\mct) = \operatorname{length}(\mct) \ge0. \] Thus a line bundle of negative degree cannot have a nonzero section.
3. If \(E\) is a divisor supported in \(X_{\reg}\) with \(\deg E=1\), then\[ \ell(E)=1. \]
Exercise IV.1.9 gives \[ \ell(E)-\ell(K-E) = \deg E+1-p_a(X) = 1. \] By step 1, \[ \deg(K-E)=-1. \] Hence step 2 gives \(\ell(K-E)=0\), and therefore \(\ell(E)=1\).
4. The assignment\[ \Phi:X_{\reg}\longrightarrow\Pic^0X, \qquad P\longmapsto[\mcl(P-P_0)] \]is well defined.
The divisor \(P-P_0\) has degree zero. By Exercise IV.1.9(a), \[ \deg\mcl(P-P_0)=0. \] Thus \(\mcl(P-P_0)\) lies in \(\Pic^0X\).
5. The map \(\Phi\) is injective.
Suppose \[ \mcl(P-P_0)\cong\mcl(Q-P_0) \] for \(P,Q\in X_{\reg}\). Tensoring by \(\mcl(P_0)\) gives \[ \mcl(P)\cong\mcl(Q). \] Thus \(P\) and \(Q\) are linearly equivalent effective divisors.
By step 3, \[ \ell(P)=1, \] so the complete linear system \(\abs{P}\) has dimension zero and contains only the divisor \(P\). Since \(Q\in\abs{P}\), we obtain \(Q=P\).
6. The map \(\Phi\) is surjective.
Let \([\mcl]\in\Pic^0X\). By Exercise IV.1.9(c), choose a divisor \(D\) supported in \(X_{\reg}\) such that \[ \mcl\cong\mcl(D). \] Since \(\mcl\) has degree zero, Exercise IV.1.9(a) gives \[ \deg D=0. \] Put \[ E=D+P_0. \] Then \(E\) is supported in \(X_{\reg}\) and has degree one. Step 3 gives \[ \ell(E)=1, \] so choose a nonzero section \[ s\in H^0(X,\mcl(E)). \]
Because \(X\) is integral, \(s\) defines an effective Cartier divisor \(Z\) and an exact sequence \[ 0 \longrightarrow \mco_X \xrightarrow{s} \mcl(E) \longrightarrow \mco_Z(E) \longrightarrow 0. \] By Exercise IV.1.9(a), \[ \deg\mcl(E)=\deg E=1. \] Taking Euler characteristics in the displayed sequence therefore gives \[ \operatorname{length}(Z)=1. \]
Hence \(Z\) is supported at a single closed point \(P\). Since \(k\) is algebraically closed, the length-one local ring of \(Z\) is \(k\). Locally at \(P\), the Cartier divisor \(Z\) is cut out by one nonzerodivisor \(a\) in the one-dimensional local domain \(\mco_{X,P}\), and \[ \mco_{X,P}/(a)\cong k. \] Thus the maximal ideal of \(\mco_{X,P}\) is generated by \(a\). A one-dimensional noetherian local ring with principal maximal ideal is regular, so \[ P\in X_{\reg}. \] Consequently \(Z=P\) as an effective Cartier divisor, and \[ \mcl(E)\cong\mcl(P). \] Since \(E=D+P_0\), \[ \mcl \cong \mcl(D) \cong \mcl(P-P_0) = \Phi(P). \] Thus \(\Phi\) is surjective.
7. Therefore\[ \boxed{ X_{\reg}\xrightarrow{\sim}\Pic^0X, \qquad P\longmapsto\mcl(P-P_0). } \]
Step 4 makes the map well defined, step 5 proves injectivity, and step 6 proves surjectivity.
8. Q.E.D. for the corrected statement.
Step 7 is the required one-to-one correspondence with \(\Pic^0X\).