Elliptic curves in 3-space

Posted on June 10, 2019

Suppose that we have an elliptic curve E/k with Weierstrass coordinates x,y. I.e., x,y\in k(E) such that (x,y,1):E\to\mathbb{P}^2 defines an isomorphism of E with the curve given by      Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ^2+a_6Z^3, with each a_i\in k and distinguished point O=[0,1,0]. Let us from now on identify E with the isomorphic Weierstrass plane curve and let x=X/Z, y=Y/Z be explicit Weierstrass coordinates for E.

The map \phi:E\to\mathbb{P}^3 given by (1,x,y,x^2) clearly maps E into the curve given by the intersection of      V_1: T_2^2+a_1T_1T_2+a_3T_2T_0=T_1T_3+a_2T_0T_3+a_4T_0T_1+a_6T_0^2 and      V_2: T_0T_3=T_1^2 in \mathbb{P}^3. Observe that this map is clearly bijectve on the the affine patch T_0\ne 0 as we then dehomogenize coordinates with T_0=1 and find T_3=T_1^2. We also find that there is a single point O&39;=[0,0,0,1] in V_1\cap V_2 satisfying T_0=0. Observe that X/Y,Z/Y generate \mathfrak{m}_{E,O} and      \frac{Z}{Y}=\left(\frac{X}{Y}\right)^3     \left(\frac{Y^2}{Y^2+a_1XY+a_3YZ-a_2X^2-a_4XZ-a_6Z^2}\right). Therefore X/Y is a local parameter at O and Z/Y has a order 3 at O. Observe that x=X/Z=(X/Y)(Y/Z) and thus x has a pole of order 2 at O; thus x^2 has a pole of order 4. Similarly we see that y=Y/Z has a pole of order 3 at O. Therefore O\mapsto [0,0,0,1]=O&39;.

Finally, observe that the map \psi:V_1\cap V_2\to E given by (T_1/T_0,T_2/T_0,1) is a rational inverse that is regular on the affine patch T_0\ne 0. It remains to check the one point O&39; in V_1\cap V_2 satisfying T_0=0. At O&39; we find, via a similar argument to the one given above for the curve E, that T_1/T_4 is a local parameter for V_1\cap V_2 at O&39;, T_2/T_4 has order 2 at O&39;, and T_0/T_4 has order 4. Therefore T_1/T_0 has a pole of order 3 at O&39; and T_2/T_0 has a pole of order 2 at O&39;. Thus \psi is regular at O&39; and O&39;\mapsto O. Hence \psi is in fact a regular inverse to \phi and E is isomorphic to V_1\cap V_2. In particular, \phi defines an isogeny between the elliptic curve E with base point O and the elliptic curve V_1\cap V_2 with base point O&39;.

Recall from Bezout’s Theorem in projective space that if F_1,\ldots,F_n\in k[X_1,\ldots,X_{n+1}]_h define hypersurfaces Q_1,\ldots,Q_n in \mathbb{P}^n in general position, i.e. \text{dim}\left(\bigcap_{i=1}^n Q_i\right)=0, then      \#\left(\bigcap_{i=1}^n Q_i\right)     =\prod_{i=1}^n\text{deg}(F_i), counting points with multiplicity. Therefore if H\subset\mathbb{P}^n is a hyperplane, the above result tells us that \#(H\cap V_1\cap V_2)=4. We made the observation above that the hyperplane defined by T_0=0 intersects V_1\cap V_2 at the point O&39; with multiplicity 4.

Suppose that P,Q,R,S\in E such that P+Q+R+S=O. Then \phi(P)+\phi(Q)+\phi(R)+\phi(S)=O&39;. In \text{Div}^0(V_1\cap V_2) this implies      (\phi(P))+(\phi(Q))+(\phi(R))+(\phi(S))-4(O&39;)=\text{div}(f) for some f\in \overline{k}(V_1\cap V_2). Observe that f is regular on the affine patch T_0\ne 0 since it has only a single pole at O&39;. Thus there exists a polynomial F\in \overline{k}[T_1,T_2,T_3] which defines a hypersurface intersecting V_1\cap V_2 at \phi(P),\phi(Q),\phi(R),\phi(S). Homogenizing we get a form F^h\in \overline{k}[T_0,T_1,T_2,T_3]_h such that F^h/T_0^{\text{deg}(F^h)}\in\overline{k}(\mathbb{P}^3) and F^h/T_0^{\text{deg}(F^h)} restricts to f\in\overline{k}(V_1\cap V_2).  \xymatrix{     \overline{k}[\mathbb{A}^3] \ar[r] & \overline{k}(\mathbb{P}^3) \ar[d]\\     \overline{k}[V_1\cap V_2\cap\mathbb{A}_3] \ar[u] \ar[r]         & \overline{k}(V_1\cap V_2) } Since f has order -4 at O&39; and T_0 intersects V_1\cap V_2 with multiplicity 4, this implies \text{deg}(F^h)=1. Thus F^h defines a hyperplane in \mathbb{P}^3 that intersects V_1\cap V_2 at \phi(P),\phi(Q),\phi(R),\phi(S).

Conversely, if H\subset\mathbb{P}^3 is a hyperplane defined by a form F that intersects V_1\cap V_2 at points \phi(P),\phi(Q),\phi(R),\phi(S), then clearly f=F/T_0\in \overline{k}(V_1\cap V_2) such that the divisor of f is (\phi(P))+(\phi(Q))+(\phi(R))+(\phi(S))-(O&39;). Thus P+Q+R+S=O.

Thus four points add to O on E if and only if there is a hyperplane in \mathbb{P}^3 intersecting \phi(E)=V_1\cap V_2 at their images under \phi. The idea for this post comes from Silverman (1, Exercise 3.10).

References

  1. Joseph Silverman, The arithmetic of elliptic curves, 2nd ed., Springer Graduate Texts in Mathematics, 1992