Let \(k\) be an algebraically closed field of characteristic zero, and let \[ X=V\bigl(y^2-x(x^2-1)\bigr)\subseteq\AA^2_k. \] Compute the divisor class group \(\Cl(X)\).
Solution
Let \(\overline X\subseteq\PP^2_k\) be the projective closure of \(X\).
1. The curve \(\overline X\) is a smooth plane cubic of genus \(1\), and\[ \overline X\setminus X=\{O\}, \qquad O=[0:1:0]. \]
In homogeneous coordinates \([X:Y:Z]\), \[ \overline X = V(F), \qquad F=Y^2Z-X^3+XZ^2. \] On the line at infinity \(Z=0\), the equation is \(X^3=0\), so the only point at infinity is \[ O=[0:1:0]. \] Moreover, \[ F_Z=Y^2+2XZ, \] so \[ F_Z(O)=1, \] and \(O\) is smooth.
On the affine chart \(Z=1\), a singular point would satisfy \[ y^2-x^3+x=0, \qquad 2y=0, \qquad 1-3x^2=0. \] The second equation gives \(y=0\), and the first then gives \[ x\in\{0,1,-1\}. \] At these three values, \[ 1-3x^2\in\{1,-2\}, \] so the third equation cannot hold. Thus \(\overline X\) is smooth everywhere.
A smooth plane cubic has genus \[ \frac{(3-1)(3-2)}2=1. \] Together with the chosen point \(O\), this makes \(\overline X\) an elliptic curve.
2. Since \(X\) is smooth,\[ \Cl(X)=\Pic(X). \]
Every local ring of a smooth curve is a discrete valuation ring. Hence every Weil divisor on \(X\) is locally principal, so every Weil divisor is Cartier. Therefore the Weil divisor class group and the Picard group coincide.
3. Restriction of divisors induces an isomorphism\[ \Pic(\overline X)\big/\ZZ[O] \xrightarrow{\sim} \Pic(X). \]
Every divisor on \(X\) is also a divisor on \(\overline X\) whose support avoids \(O\), so restriction gives a surjection \[ \Pic(\overline X)\longrightarrow\Pic(X). \]
Suppose a divisor class \([D]\in\Pic(\overline X)\) restricts to zero in \(\Pic(X)\). Then for some \[ g\in k(X)^\times=k(\overline X)^\times \] one has \[ D|_X=\div_X(g). \] Consequently, \[ D-\div_{\overline X}(g) \] is supported on the single missing point \(O\), so it equals \(mO\) for some \(m\in\ZZ\). Thus \[ [D]\in\ZZ[O]. \] Conversely, the divisor \(O\) restricts to the zero divisor on \(X\). Hence the kernel is exactly \(\ZZ[O]\), proving the quotient description.
4. There is a canonical isomorphism\[ \Pic(\overline X)\big/\ZZ[O] \xrightarrow{\sim} \Pic^0(\overline X). \]
The degree homomorphism \[ \deg:\Pic(\overline X)\longrightarrow\ZZ \] has kernel \(\Pic^0(\overline X)\), and the map \[ n\longmapsto[nO] \] is a section because \(\deg(nO)=n\). Therefore \[ \Pic(\overline X) \cong \Pic^0(\overline X)\oplus\ZZ[O]. \] Quotienting by the second summand gives the displayed isomorphism.
5. The Abel–Jacobi map\[ \alpha:\overline X(k)\longrightarrow\Pic^0(\overline X), \qquad P\longmapsto[P-O], \]is an isomorphism of groups.
Let \(D\) be a divisor of degree \(0\) on \(\overline X\). Since \(\overline X\) has genus \(1\), Riemann–Roch applied to the degree-one divisor \(D+O\) gives \[ \ell(D+O)=1. \] Hence there is a unique effective divisor of degree \(1\) in the class of \(D+O\). Such a divisor is a single point \(P\), so \[ D\sim P-O. \] This proves surjectivity of \(\alpha\).
If \[ P-O\sim Q-O, \] then \(P\) and \(Q\) are effective divisors in the same degree-one complete linear system. The uniqueness just proved gives \(P=Q\), so \(\alpha\) is injective.
Under the usual elliptic-curve group law with identity \(O\), this bijection is a group isomorphism (the elliptic-curve group law).
6. The divisor class group of the affine curve is\[ \boxed{ \Cl(X) \cong \overline X(k), } \]where \(\overline X(k)\) carries its elliptic-curve group law with identity\(O=[0:1:0]\).
Combining steps <1>2–<1>5 gives \[ \Cl(X) \cong \Pic(X) \cong \Pic(\overline X)/\ZZ[O] \cong \Pic^0(\overline X) \cong \overline X(k). \]
7. Equivalently, every divisor on \(X\) is linearly equivalent either to\(0\) or to a point \(P\in X(k)\).
Under the isomorphism of step 6, the identity element \(O\in\overline X(k)\) corresponds to the zero class because the divisor \(O\) disappears on \(X=\overline X\setminus\{O\}\).
Every other point of \(\overline X(k)\) lies in \(X(k)\). For \[ P\in X(k), \] the degree-zero class \([P-O]\) on \(\overline X\) restricts to the divisor class \([P]\) on \(X\). Hence every nonzero class of \(\Cl(X)\) is represented by a point of \(X\).
8. Q.E.D.
Step 6 computes the group, and step 7 represents each class by \(0\) or a point of \(X\).