Elliptic curves in 3-space

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 $(1,x,y,x^2)$ 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 bijectve on 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 $f\in\overline{k}(V_1\cap V_2).$ $\xymatrix{ \overline{k}[\mathbb{A}^3] \ar[r]\ar[d] & \overline{k}(\mathbb{P}^3) \ar[d]\\ \overline{k}[V_1\cap V_2\cap\mathbb{A}_3] \ar[r] & \overline{k}(V_1\cap V_2) }$ 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 $(\phi(P))+(\phi(Q))+(\phi(R))+(\phi(S))-4(O&39;).$ 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

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