Let \(X\) be an elliptic curve in \(\PP^2\) given by an equation of the form \[ y^2+a_1 x y+a_3 y=x^3+a_2 x^2+a_4 x+a_6 . \] Show that the \(j\)-invariant is a rational function of the \(a_i\), with coefficients in \(\QQ\). In particular, if the \(a_i\) are all in some field \(k_0 \subseteq k\), then \(j \in k_0\) also. Furthermore, for every \(\alpha \in k_0\), there exists an elliptic curve defined over \(k_0\) with \(j\)-invariant equal to \(\alpha\).
Solution
Throughout this section Hartshorne assumes \(\characteristic k\ne2\).
1. After completing the square, the equation becomes\[ Y^2=x^3+Ax^2+Bx+C, \]where\[ A=a_2+\frac{a_1^2}{4},\qquad B=a_4+\frac{a_1a_3}{2},\qquad C=a_6+\frac{a_3^2}{4}. \]
Set \[ Y=y+\frac{a_1x+a_3}{2}. \] Expanding \(Y^2\) and substituting the original equation gives the displayed cubic. This coordinate change is defined because \(\characteristic k\ne2\) and does not change the isomorphism class, hence not the \(j\)-invariant.
2. Let \(p(x)=x^3+Ax^2+Bx+C\) and let\[ \Delta_p=A^2B^2-4B^3-4A^3C-27C^2+18ABC \]be its polynomial discriminant. Then\[ \boxed{j(X)=256\frac{(A^2-3B)^3}{\Delta_p}}. \]
Write \[ p(x)=(x-r_1)(x-r_2)(x-r_3). \] The roots are distinct: otherwise a repeated root \(r\) would make \((r,0)\) singular on \(Y^2=p(x)\).
Set \[ x=r_1+(r_2-r_1)X. \] Then the three roots become \(0,1,\lambda\), where \[ \lambda=\frac{r_3-r_1}{r_2-r_1}. \] The right-hand side acquires the nonzero factor \((r_2-r_1)^3\). Since \(k\) is algebraically closed, choose \(s\in k^\times\) with \(s^2=(r_2-r_1)^3\) and set \(Y=sY'\). Thus the curve is isomorphic to the Legendre curve \[ Y'^2=X(X-1)(X-\lambda). \] Hartshorne’s Legendre formula is therefore \[ j=256\frac{(\lambda^2-\lambda+1)^3} {\lambda^2(\lambda-1)^2}. \] By Vieta, \[ r_1+r_2+r_3=-A, \qquad r_1r_2+r_1r_3+r_2r_3=B, \] and direct simplification gives \[ \lambda^2-\lambda+1 =\frac{A^2-3B}{(r_2-r_1)^2}. \] Also \[ \Delta_p=(r_1-r_2)^2(r_1-r_3)^2(r_2-r_3)^2. \] Substitution in the Legendre formula cancels the powers of \(r_2-r_1\) and gives the claimed expression. Expanding the symmetric product of squared root differences gives the displayed polynomial formula for \(\Delta_p\).
3. Hence \(j\) is a rational function of the \(a_i\) with coefficients in\(\QQ\), and if all \(a_i\) lie in \(k_0\), then \(j\in k_0\).
Step 1 expresses \(A,B,C\) polynomially in the \(a_i\) with rational coefficients, and step 2 expresses \(j\) rationally in \(A,B,C\) with integer coefficients. Since \(X\) is nonsingular, \(\Delta_p\ne0\).
If the \(a_i\) lie in \(k_0\), then \(\characteristic k_0\ne2\), so \(1/2\) and \(1/4\) belong to \(k_0\). Thus \(A,B,C,\Delta_p\) and consequently \(j\) all lie in \(k_0\).
4. Suppose \(\characteristic k_0\ne2,3\). Then every\(\alpha\in k_0\) occurs as the \(j\)-invariant of an elliptic curve definedover \(k_0\).
For \(\alpha=0\), take \[ E_0:y^2=x^3+1. \] Its cubic discriminant is \(-27\ne0\), and step 2 gives \(j(E_0)=0\).
For \(\alpha=1728\), take \[ E_{1728}:y^2=x^3-x. \] Its cubic discriminant is \(4\ne0\), and step 2 gives \(j(E_{1728})=1728\).
Now assume \(\alpha\ne0,1728\) and put \[ u=-\frac{3\alpha}{\alpha-1728}, \qquad v=-\frac{2\alpha}{\alpha-1728}. \] Take \[ E_\alpha:y^2=x^3+ux+v. \] For a cubic \(x^3+ux+v\), step 2 becomes \[ j=1728\frac{4u^3}{4u^3+27v^2}. \] Here \[ 4u^3+27v^2 =- \frac{108\cdot1728\,\alpha^2}{(\alpha-1728)^3} \ne0, \] so the curve is nonsingular, and direct substitution gives \[ j(E_\alpha)=\alpha. \] All coefficients belong to \(k_0\).
5. Suppose \(\characteristic k_0=3\). Again every\(\alpha\in k_0\) occurs over \(k_0\).
Here \(1728=0\). For \(\alpha=0\), take \[ E_0:y^2=x^3-x. \] Its cubic discriminant is \(4=1\), while \(A^2-3B=0\), so \[ j(E_0)=0. \]
If \(\alpha\ne0\), take \[ E_\alpha:y^2=x^3+x^2-\alpha^{-1}. \] For this cubic, \[ A=1,\qquad B=0,\qquad C=-\alpha^{-1}. \] Hence in characteristic \(3\), \[ A^2-3B=1, \qquad \Delta_p=4\alpha^{-1}=\alpha^{-1}\ne0. \] Therefore \(E_\alpha\) is nonsingular and \[ j(E_\alpha) =256\frac1{\alpha^{-1}} =\alpha, \] because \(256=1\) in characteristic \(3\).
6. Q.E.D.
Steps <1>1–<1>3 give the rational expression and descent of \(j\) to the coefficient field. Steps <1>4–<1>5 give an elliptic curve over \(k_0\) with every prescribed \(j\)-invariant \(\alpha\in k_0\).