Tri-homogenous polynomials of tridegree $(3,3,3)$ to add three points on an elliptic curve

Question

Consider an elliptic curve $E \subset \mathbb{P}^2$ with the zero point $\mathcal{O}$. There are classical articles about complete systems of addition laws on $E$ (see
Lange and Ruppert - Complete systems of addition laws on abelian varieties,
Bosma - Complete systems of two addition laws for elliptic curves). It is known that the group law on $E$ can be represented by bi-homogenous polynomials of bidegree $(2, 2)$. More precisely, those articles show that for the divisor $D := 3(2V + 2H -\triangledown)$ the global section space $H^{0}(E^2, D)$ is non-zero, where $\triangledown$ is the anti-diagonal, $V := E \times \{\mathcal{O}\}$, and $H := \{\mathcal{O}\} \times E$. Moreover, $\lvert D\rvert$ is a base-point free linear system.

I am interested in the addition of three points $P := P_0 + P_1 + P_2$ on $E$. Of course, $P = (P_0 + P_1) + P_2$. However, expressing $P$ in this way gives rise to tri-homogenous polynomials of tridegree $(4,4,2)$. Alternatively, I guess that there are tri-homogenous polynomials of tridegree $(3,3,3)$. Since $3\cdot 3 = 9 < 10 = 2\cdot 4+2$, it seems that the latter polynomials can be evaluated more efficiently. This may have applications in elliptic cryptography.

Now let $D := 3(3A_0 + 3A_1 + 3A_2 - B)$, where $$ A_0 := \{\mathcal{O}\} \times E^2, \quad A_1 := E \times \{\mathcal{O}\} \times E, \quad A_2 := E^2 \times \{\mathcal{O}\}, \quad B := \big\{(P_0, P_1, -P_0-P_1)\big\} \quad \subset \quad E^3. $$ If I am not mistaken, to confirm my conjecture it is necessary to prove that the space $H^{0}(E^3, D)$ is non-zero. Help me please to do this. And what about base-point freeness of $\lvert D\rvert$?

Answer 1

The divisor $D$ is rationally equivalent to an effective divisor, hence $H^0(E^3,D) \neq 0$.

To see this, let $p_0,p_1,p_2 \colon E^3 \to E$ be the canonical projections. For $i,j \in \{0,1,2\}$, write $p_{i,j}=p_i+p_j$. By the theorem of the cube [1, Corollary 6.4], $B$ is rationally equivalent to \begin{equation*} B \sim p_{0,1}^* (0) + p_{0,2}^* (0) + p_{1,2}^* (0) - A_0 - A_1 - A_2. \end{equation*} Moreover, on $E \times E$ we have the relation $\Delta + \nabla - 2H -2V \sim 0$, where $\Delta$ is the diagonal, $\nabla = \{(x,y) \in E \times E: x+y=0\}$ is the anti-diagonal, $H = E \times \{0\}$ and $V = \{0\} \times E$. Therefore \begin{align*} D & \sim 3 (3A_0+3A_1+3A_2 - p_{0,1}^*(0) - p_{0,2}^*(0) - p_{1,2}^*(0) +A_0+A_1+A_2) \\ & \sim 3 \bigl(4A_0+4A_1+4A_2 - (2A_1+2A_0-\Delta_{0,1}) - (2A_2+2A_0-\Delta_{0,2}) - (2A_2+2A_1-\Delta_{1,2})\bigr) \\ & = 3 (\Delta_{0,1} + \Delta_{0,2} + \Delta_{1,2}), \end{align*} where $\Delta_{i,j} = p_{i,j}^* \Delta$ is the pull-back of the diagonal under the projection $p_{i,j} \colon E^3 \to E^2$ onto the factors with indices $i,j$. This proves the claim.

[1] Milne, J. S. Abelian varieties. Arithmetic geometry (Storrs, Conn., 1984), 103–150, Springer, New York, 1986.