Schoof's algorithm: Difference between revisions

Revision as of 12:20, 10 May 2018

Revision as of 20:20, 10 May 2018

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We must split the problem into two cases: the case in which <math>(x^{q^{2}}, y^{q^{2}}) \neq \pm \bar{q}(x, y)</math>, and the case in which <math>(x^{q^{2}}, y^{q^{2}}) = \pm \bar{q}(x, y)</math>. Note that these equalities are checked modulo <math>\psi_l</math>.

===Case 1: <math>(x^{q^{2===}}, y^{q^{2}}) \neq \pm \bar{q}(x, y)</math>}}===

By using the [[Elliptic curves#The group law|addition formula]] for the group <math>E(\mathbb{F}_{q})</math> we obtain:

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As mentioned earlier, using {{mvar|Y}} and <math>y_{\bar{t}}^{q}</math> we are now able to determine which of the two values of <math>\bar{t}</math> (<math>\bar{t}</math> or <math>-\bar{t}</math>) works. This gives the value of <math>t\equiv \bar{t}\pmod l</math>. Schoof's algorithm stores the values of <math>\bar{t}\pmod l</math> in a variable <math>t_l</math> for each prime {{mvar|l}} considered.

===Case 2: <math>(x^{q^{2===}}, y^{q^{2}}) = \pm \bar{q}(x, y)</math>}}===

We begin with the assumption that <math>(x^{q^{2}}, y^{q^{2}}) = \bar{q}(x, y)</math>. Since {{mvar|l}} is an odd prime it cannot be that <math>\bar{q}(x, y)=-\bar{q}(x, y)</math> and thus <math>\bar{t}\neq 0</math>. The characteristic equation yields that <math>\bar{t} \phi(P) = 2\bar{q} P</math>. And consequently that <math>\bar{t}^{2}\bar{q} \equiv (2q)^{2} \pmod l</math>.

This implies that {{mvar|q}} is a square modulo {{mvar|l}}. Let <math>q \equiv w^{2} \pmod l</math>. Compute <math>w\phi(x,y)</math> in <math>\mathbb{F}_{q}[x,y]/(y^{2}-x^{3}-Ax-B, \psi_{l})</math> and check whether <math>