Let \(X\) be the curve \(y^2+y=x^3-x\) of (4.23.8).
If \(Q=(a, b)\) is a point on the curve, compute the coordinates of the point \(P+Q\), where \(P=(0,0)\), as a function of \(a, b\). Use this formula to find the coordinates of \(nP\), \(n=1,2, \ldots, 10\). Check: \(6P = (6,14)\).
This equation defines a nonsingular curve over \(\FF_p\) for all \(p \neq 37\).
Solution
Write \[ y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 \] with \[ a_1=a_2=a_6=0, \qquad a_3=1, \qquad a_4=-1. \]
1. If \(Q=(a,b)\) and \(a\ne0\), then\[ \boxed{ P+Q= \left( \frac{b^2}{a^2}-a, -\frac ba\left(\frac{b^2}{a^2}-a\right)-1 \right). } \]
The line through \(P=(0,0)\) and \(Q=(a,b)\) is \(y=\lambda x\) with \(\lambda=b/a\). Substitution gives \[ x\bigl(x^2-\lambda^2x-(\lambda+1)\bigr)=0. \] If \(R\) is the third intersection point, Vieta gives \[ x(R)=\lambda^2-a, \qquad y(R)=\lambda x(R). \] On this Weierstrass equation \[ -(x,y)=(x,-y-1). \] Since \(P+Q=-R\), the displayed formula follows.
2. The points with first coordinate \(0\) are \(P=(0,0)\) and\(-P=(0,-1)\), and\[ \boxed{2P=(1,0)}. \]
At \(x=0\) the equation is \(y(y+1)=0\). For the tangent computation, put \[ G(x,y)=y^2+y-x^3+x. \] Since \(G_x(P)=G_y(P)=1\), the tangent at \(P\) is \(y=-x\). Its third intersection with \(X\) is \((1,-1)\), whose inverse is \((1,0)\).
3. The first ten multiples of \(P\) are\[ \boxed{ \begin{array}{c|c} n & nP\\ \hline 1&(0,0)\\ 2&(1,0)\\ 3&(-1,-1)\\ 4&(2,-3)\\ 5&\left(\frac14,-\frac58\right)\\ 6&(6,14)\\ 7&\left(-\frac59,\frac8{27}\right)\\ 8&\left(\frac{21}{25},-\frac{69}{125}\right)\\ 9&\left(-\frac{20}{49},-\frac{435}{343}\right)\\ 10&\left(\frac{161}{16},-\frac{2065}{64}\right). \end{array} } \]
Step 2 gives the first two entries. Applying step 1 successively gives \[ \begin{aligned} P+(1,0)&=(-1,-1),\\ P+(-1,-1)&=(2,-3),\\ P+(2,-3)&=\left(\frac14,-\frac58\right),\\ P+\left(\frac14,-\frac58\right)&=(6,14),\\ P+(6,14)&=\left(-\frac59,\frac8{27}\right),\\ P+\left(-\frac59,\frac8{27}\right) &=\left(\frac{21}{25},-\frac{69}{125}\right),\\ P+\left(\frac{21}{25},-\frac{69}{125}\right) &=\left(-\frac{20}{49},-\frac{435}{343}\right),\\ P+\left(-\frac{20}{49},-\frac{435}{343}\right) &=\left(\frac{161}{16},-\frac{2065}{64}\right). \end{aligned} \] In particular \(6P=(6,14)\), as required.
4. The discriminant of the generalized Weierstrass equation is\[ \boxed{\Delta=37.} \]
For the coefficients \[ a_1=a_2=a_6=0, \qquad a_3=1, \qquad a_4=-1, \] the standard quantities are \[ b_2=0, \qquad b_4=-2, \qquad b_6=1, \qquad b_8=-1. \] Therefore \[ \begin{aligned} \Delta &=-b_2^2b_8-8b_4^3-27b_6^2+9b_2b_4b_6\\ &=-8(-2)^3-27\\ &=37. \end{aligned} \]
5. For every prime \(p\ne37\), the reduction of\[ y^2+y=x^3-x \]over \(\FF_p\) is nonsingular.
A generalized Weierstrass cubic is nonsingular over a field exactly when its discriminant is nonzero in that field. By step 4, the discriminant of this integral model reduces to the class of \(37\) in \(\FF_p\). Thus it is nonzero for every \(p\ne37\), proving part (b).
6. Q.E.D.
Steps <1>1–<1>3 prove part (a), and steps <1>4–<1>5 prove part (b).