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](https://doi.org/10.1007/BF01388526),  

[Bosma - Complete systems of two addition laws for elliptic curves](https://www.math.ru.nl/%7Ebosma/pubs/JNT1995.pdf)). 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 $\lvert D\rvert$?



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


