Define the Weierstrass cubic as \(X\da V(y^{2} z-(x^{3}+g_{2} x z^{2}+g_{3} z^{3}) )\) for \(g_2, g_3 \in \CC\). Show that \(X\) is nonsingular iff \(p(x) \da x^3 + g_2 x + g_3\) has no multiple roots.
Solution
Put \[ F(x,y,z) = y^2z-x^3-g_2xz^2-g_3z^3 \] and \[ p(x)=x^3+g_2x+g_3. \]
1. The only point of \(X\) on the line \(z=0\) is\[ O=(0:1:0), \]and \(O\) is nonsingular.
On \(z=0\) the equation \(F=0\) becomes \[ -x^3=0, \] so \(x=0\). Since \((x:y:z)\) is a projective point, \(y\ne0\), and hence the only such point is \(O=(0:1:0)\).
The partial derivative with respect to \(z\) is \[ F_z = y^2-2g_2xz-3g_3z^2. \] Therefore \[ F_z(O)=1\ne0. \] By the projective Jacobian criterion, \(O\) is nonsingular.
2. A point \((a:b:1)\in X\) is singular if and only if\[ b=0, \qquad p(a)=0, \qquad p'(a)=0. \]
On the affine chart \(z=1\), the curve is \[ f(x,y) = y^2-p(x) = 0. \] Its partial derivatives are \[ f_x=-p'(x)=-(3x^2+g_2), \qquad f_y=2y. \] Since the ground field is \(\CC\), the affine Jacobian criterion says that \((a,b)\) is singular exactly when \[ f(a,b)=f_x(a,b)=f_y(a,b)=0. \] The equation \(f_y(a,b)=0\) is equivalent to \(b=0\). With \(b=0\), the remaining two equations are precisely \[ p(a)=0 \qquad\text{and}\qquad p'(a)=0. \] This proves the claim.
3. The curve \(X\) is singular if and only if \(p\) has a multiple root.
By step 1, the point at infinity is nonsingular, so every singular point of \(X\) lies in the chart \(z=1\). By step 2, such a singular point exists exactly when there is some \(a\in\CC\) satisfying \[ p(a)=p'(a)=0. \] A root of a polynomial over a field of characteristic zero is multiple if and only if it is also a root of the derivative. Thus the displayed condition is equivalent to \(p\) having a multiple root.
4. Equivalently,\[ X\text{ is nonsingular} \quad\Longleftrightarrow\quad 4g_2^3+27g_3^2\ne0. \]
Suppose first that \(a\) is a multiple root of \(p\). Then \[ 3a^2+g_2=0 \qquad\text{and}\qquad a^3+g_2a+g_3=0. \] The first equation gives \(g_2=-3a^2\), and substituting this into the second gives \(g_3=2a^3\). Hence \[ 4g_2^3+27g_3^2 = 4(-3a^2)^3+27(2a^3)^2 = 0. \]
Conversely, suppose \[ 4g_2^3+27g_3^2=0. \] If \(g_2=0\), then \(g_3=0\), and \(a=0\) is a common root of \(p\) and \(p'\). If \(g_2\ne0\), put \[ a=-\frac{3g_3}{2g_2}. \] The displayed relation gives \[ a^2 = \frac{9g_3^2}{4g_2^2} = -\frac{g_2}{3}, \] so \(p'(a)=3a^2+g_2=0\). Moreover, \[ \begin{aligned} p(a) &=a(a^2+g_2)+g_3\\ &=\frac23ag_2+g_3\\ &=0. \end{aligned} \] Thus \(p\) has a multiple root. Therefore its discriminant \[ -4g_2^3-27g_3^2 \] is nonzero exactly when \(X\) is nonsingular, by step 3.
5. Q.E.D.
Step 3 proves the required equivalence, and step 4 records it in the usual discriminant form.