Let \[ C\subseteq\PP^2_k \] be a smooth plane cubic over the algebraically closed ground field \(k\). Show that \(C\) is an elliptic curve.
Solution
1. The smooth plane cubic \(C\) is irreducible.
Let \[ C=V(F), \] where \(F\) is a homogeneous cubic. Suppose that \(F\) were reducible. Then \[ F=GH \] with \(G\) and \(H\) nonconstant homogeneous polynomials of positive degree.
Over the algebraically closed field \(k\), the projective plane curves \[ V(G) \qquad\text{and}\qquad V(H) \] have a common point \(P\) by Bézout’s theorem. At such a point, \[ G(P)=H(P)=0. \] For every homogeneous coordinate \(x_i\), \[ \frac{\partial F}{\partial x_i} = H\frac{\partial G}{\partial x_i} + G\frac{\partial H}{\partial x_i}, \] so \[ \frac{\partial F}{\partial x_i}(P)=0. \] Thus \(P\) is singular on \(C\), contradicting smoothness. Therefore \(F\) is irreducible and \(C\) is an integral projective curve.
2. The geometric genus of \(C\) is\[ \boxed{g(C)=1.} \]
For an integral plane curve of degree \(d\), the genus formula D-CRVPLSING gives \[ g = \binom{d-1}{2} - \sum_{p\in\operatorname{Sing}(C)}\delta_p. \] Here \[ d=3 \] and \(C\) is smooth, so the singularity sum is empty. Hence \[ g(C) = \binom{2}{2} = 1. \]
3. The curve \(C\) is elliptic.
An elliptic curve is a projective curve of geometric genus \(1\), and step 2 gives this condition. Therefore \[ \boxed{C\text{ is an elliptic curve}.} \]
4. Q.E.D.
Step 1 verifies that the smooth cubic is an integral projective curve, step 2 computes its genus, and step 3 concludes that it is an elliptic curve.