For a finite field $F_{p^2}$, polynomial elements appear. For example, $F_{2^2}=\{0,1,x,x+1\}$.

Now when defining an Elliptic Curve $E:y^2=x^3+Ax+B$ over a field of the form $F_{p^2}$.

Does this mean that $A$ and $B$ can be polynomials??

  • 2
    $\begingroup$ Strictly speaking, $A$ and $B$ are elements of the field $F_{2^2}$ that we are realizing as a quotient $F_2[t] / \langle p(t) \rangle$ of the polynomial ring $F_2[t]$ by the (maximal) ideal generated by a particular irreducible polynomial (in this case our only choice is $p(t) = t^2 + t + 1$). Alternatively, we can regard elements of the field as elements generated over $F_2$ by $\alpha$ and subject to the relation $\alpha^2 + \alpha + 1 = 0$. In either case, it's misleading to use the coordinate $x$ appearing in the definition of the elliptic curve for the indeterminate/adjoined element. $\endgroup$ – Travis Willse Mar 20 '17 at 14:55
  • $\begingroup$ So if $E:y^2=x^3+Ax+B$ over $F_{2^2}=\{0,1,α,α+1\}$ then, $y^2=x^3+αx+α+1$ is an elliptic curve defined over $F_{2^2}$?? And also, what is '$α$'. Is it a constant? I understand that it's an indetermined element, but are there ways to determine it? $\endgroup$ – Junsworth Mar 20 '17 at 15:03
  • $\begingroup$ Yes, that's correct. If we think of $F_{2^2}$ as the quotient $F_2[t] / \langle t^2 + t + 1\rangle$, then $\alpha$ is just the image $t + \langle t^2 + t + 1 \rangle$ of $t$ under the quotient map $F_2[t] \to F_2[t] / \langle t^2 + t + 1\rangle$. $\endgroup$ – Travis Willse Mar 20 '17 at 15:14
  • $\begingroup$ Ok great, is it correct to consider $A,\; B$ as elements of the quotient ring ? $\endgroup$ – Junsworth Mar 20 '17 at 15:18
  • $\begingroup$ Yes, or more to the point, $A, B$ are elements of the field. By the way, note that because our underlying field $F_{2^2}$ has characteristic $2$, /not/ all elliptic curves over that field can be written in the usual reduced form $y^2 = x^3 + A x + B$ after applying a suitable affine transformation. The best we can do in our case is (I think) $y^2 + A x y + B y = x^3 + C x^2 + D x + E$. $\endgroup$ – Travis Willse Mar 20 '17 at 15:31

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