4
$\begingroup$

Usually, an elliptic curve is defined to be a smooth cubic curve in $\mathbb{P}^2$. But twisted Edward curves defined by $$ ax^2 + y^2 = 1 + dx^2y^2 $$ over fields of characteristics $\neq 2$, have degree 4, and they have singularities at infinity: $(0:1:0)$ and $(1:0:0)$. I know that a twisted Edwards curve is birationally equivalent to a smooth cubic curve in Montgomery form. But birational equivalence does not equal isomorphism.

Is a twisted Edwards curve including points at infinity a group and why? If it's not, which part of it forms a group and why?

$\endgroup$
1
  • $\begingroup$ You seem to doubt that birational equivalence does not allow you to define a group structure of the elliptic curve's points. The Wikipedia article Twisted Edwards curve gives a direct account of addition without reference to birational equivalence to Montgomery curves. For more about the advantages of computing with one form or the other see Twisted Edwards Curves Revisited. $\endgroup$
    – hardmath
    Commented Dec 30, 2022 at 20:31

1 Answer 1

5
$\begingroup$

You've written an affine equation for the twisted Edwards curve $C$, giving it as a subset of $\newcommand{\A}{\mathbb{A}} \newcommand{\P}{\mathbb{P}} \mathbb{A}^2$, but in mentioning the points $(0:1:0)$ and $(1:0:0)$, you are implicitly considering it as a subset of $\mathbb{P}^2$. However, there are other projective varieties naturally containing $\mathbb{A}^2$ as an open, and we can equally as well take the closure of $C$ in one of these---see the Extended coordinates section of the Wikipedia article for a few examples.

Here's a different proposal from those listed at that page: consider $C \subseteq \mathbb{A}^2 \subseteq \mathbb{P}^1 \times \mathbb{P}^1$ and take its closure in $\mathbb{P}^1 \times \mathbb{P}^1$. Writing $((X_0 : X_1), (Y_0 : Y_1))$ for the coordinates on $\mathbb{P}^1 \times \mathbb{P}^1$, then we can identify $\A^2$ as the open subset where $X_1 \neq 0$ and $Y_1 \neq 0$, with affine coordinates $x = X_0/X_1, y = Y_0/Y_1$. (There is a reason for choosing $\P^1 \times \P^1$ coming from toric geometry and the Newton polygon of the Edwards curve equation.) Substituting for $x$ and $y$ and clearing denominators, we find that the closure $\overline{C}$ of $C$ in $\P^1 \times \P^1$ is given by the bihomogeneous equation $$ \overline{C} : a X_0^2 Y_1^2 + X_1^2 Y_0^2 = X_1^2 Y_1^2 + d X_0^2 Y_0^2 \, . $$ (This equation is bihomogeneous of degree $(2,2)$, so $\overline{C}$ has genus $(2-1)(2-1) = 1$, as expected.)

Setting $X_1 = 0$, we have \begin{align*} 0 &= a X_0^2 Y_1^2 - d X_0^2 Y_0^2 = X_0^2 (a Y_1^2 - d Y_0^2) = X_0^2 \left(\sqrt{a} Y_1 - \sqrt{d} Y_0\right) \left(\sqrt{a} Y_1 + \sqrt{d} Y_0\right) \, . \end{align*} Since $X_0 \neq 0$ (we already have $X_1 = 0$), we find the two points $$ ((1:0), (\sqrt{a} : \sqrt{d})), \quad ((1:0), (\sqrt{a} : -\sqrt{d})) \, , $$ possibly defined over some extension of the base field. Setting $Y_0 = 0$ and proceeding similarly, we find two more points $$ ((1:\sqrt{d}), (1:0)), \quad ((1:-\sqrt{d}), (1:0)) $$ giving 4 total points at infinity. Moreover, one can show that $\overline{C}$ is nonsingular at each of these 4 points, so is nonsingular everywhere. (This is assuming $a \neq 0, d \neq 0$, and $a \neq d$, as required in the definition of a twisted Edwards curve.)

I claim that all the points of $\overline{C}$, including the 4 points at infinity found above, form a group. Moreover, we can obtain the group law by appropriately homogenizing the formula for the group law on the usual affine piece, given in section 6 of Bernstein et. al's Twisted Edwards Curves. The formula for $((X_0 : X_1), (Y_0 : Y_1)) + ((X_0' : X_1'), (Y_0' : Y_1'))$ is \begin{equation} \begin{aligned} \left(\left(X_0 Y_0' X_1' Y_1 + X_0' Y_0 X_1 Y_1' : X_1 X_1' Y_1 Y_1' + d X_0 X_0' Y_0 Y_0'\right),\\ \left(Y_0 Y_0' X_1 X_1' - a X_0 X_0' Y_1 Y_1' : X_1 X_1' Y_1 Y_1' - d X_0 X_0' Y_0 Y_0' \right) \right) \, . \end{aligned} \tag{$*$} \label{add} \end{equation}

This formula works even for the points at infinity. For instance, doubling the points $((1:0), (\sqrt{a} : \pm\sqrt{d}))$ using (\ref{add}), we obtain \begin{align*} 2 \cdot ((1:0), (\sqrt{a} : \pm\sqrt{d})) &= ((0 : d \cdot \sqrt{a} \cdot \sqrt{a}), (-a \cdot (\pm\sqrt{d}) \cdot (\pm \sqrt{d}) : -d \cdot \sqrt{a} \cdot \sqrt{a}))\\ &= ((0:ad), (-ad : -ad)) = ((0:1), (1:1)) \, . \end{align*} This is the identity of the group law, so $((1:0), (\sqrt{a} : \pm \sqrt{d}))$ are $2$-torsion points. Doubling $((1 : \pm \sqrt{d}), (1:0))$ using (\ref{add}), we find $$ 2 \cdot ((1 : \pm \sqrt{d}), (1:0)) = ((0:1), (1:-1)) \, . $$ Doubling $((0:1), (1:-1))$, we obtain $2 \cdot ((0:1), (1:-1)) = ((0:1), (1:1))$, showing that $((1 : \pm \sqrt{d}), (1:0))$ are $4$-torsion points. This agrees with the observations in the section Exceptional Points for the Birational Equivalence of Bernstein et al.

$\endgroup$
3
  • $\begingroup$ if your curve $\overline{C}$ is smooth projective and of genus $1$ it is an elliptic curve when $char(k)\neq 2$. Any such curve embeds into $\mathbb{P}^2_k$ - why do you choose to embed it into $(\mathbb{P}^1)^2$? $\endgroup$
    – hm2020
    Commented Jan 1, 2023 at 9:58
  • 3
    $\begingroup$ @hm2020 Edwards proposed an alternative to the Weierstrass model for elliptic curves in his paper A Normal Form for Elliptic Curves. In the Edwards model, the group law has a particularly nice form, reminiscent of the angle-addition formulae for $\sin$ and $\cos$. This form of the group law allows for faster computations than on Weierstrass models, so has become of interest to cryptographers. See the linked paper of Bernstein et al. for more on the efficiency of arithmetic on Edwards models. $\endgroup$ Commented Jan 1, 2023 at 18:32
  • 1
    $\begingroup$ - thanks for the reference. $\endgroup$
    – hm2020
    Commented Jan 1, 2023 at 19:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .