Exercise 5.7.23* ($\mathscr{C}_f(\mathbb{R}) \simeq \mathbb{R}/\mathbb{Z}$ if $\mathscr{C}_f(\mathbb{R}) $ has one connected component)

Warning: This solution is very incomplete and still being researched. It is more of a complete theory than a problem. So I will leave it as is for now, so that I can continue to advance other parts of this book.

Let $𝒞$ be the elliptic curve defined by $y^2-x^3-a{x}^2-bx-c,$ where the polynomial $x^3-a{x}^2-bx-c$ has a unique real root and two complex non real roots.

By translation on the $x$-axis, we may suppose that the coefficient of $x^2$ is zero, so that

and $p(x)=x^3+px+q$ has exactly one real root $\alpha =r$, and two non real complex roots $\beta =s-it$ and $\gamma =s+it$ ( $s,t\in ℝ$), and

Then

(a)

Division by $2$of a point on a real elliptic curve.

Let $P=(X,Y)$ be any point on $𝒞$ ( $P\ne O$). It is not obvious that the equation $2M=P$, where $M=(x,y)$ is the unknown, has two real solutions on $𝒞$, one such as $y<0$ and the other such that $y>0$. If we want to give an elementary proof (which avoids the use of the Weierstrass $\wp$-function), this requires some calculations.

If $2M=P$, then the tangent $L$ to the curve at $M$ contains $-P$. Then $L$ is not a vertical line, otherwise $M=O$, so $P=O$ which is excluded. Therefore the tangent line $L$ passing by $-P=(X,-Y)$ has an equation of the form $y+Y=m(x-X)$, that is

$$y=mx-mX-Y.$$

The abscissas of the points of intersection of this line with the curve are given by the equation

$$\begin{array}{lll}\hfill 0& ={(mx-mX-Y)}^2-x^3-px-q.& \hfill \text{(2)}\end{array}$$

Since

$$\begin{array}{llll}\hfill 0& =Y^2-X^3-pX-q,& \hfill & \end{array}$$

the equation (2) is equivalent to

$$\begin{array}{llll}\hfill 0& ={(mx-mX-Y)}^2-Y^2-(x^3-X^3)-p(x-X)& \hfill & \\ \hfill & =(X-x)\cdot (x^2+\left(-m^2+X\right)x+m^{2}X+X^2+2mY+p).& \hfill & \end{array}$$

Except the point $P$, the two other intersection points are given by

$$\begin{array}{lll}\hfill 0=x^2+\left(-m^2+X\right)x+m^{2}X+X^2+2mY+p.& & \hfill \text{(3)}\end{array}$$

The line $L$ has a contact of order $2$ with the curve if and only if the discriminant of this second degree polynomial is $0$. This gives the condition for $m$ to be the slope of a tangent to the curve (except the tangent at point $P$):

$$\begin{array}{lll}\hfill 0& =\mathrm{\Delta }(m)={\left(m^2-X\right)}^2-4(m^{2}X+X^2+2mY+p).& \hfill \text{(4)}\end{array}$$

This equation has degree 4, because there are four complex solutions of the complex elliptic curve ${𝒞}_f(ℂ)$. We want to show that there are two real solutions.

This can be shown by noting that the discriminant $D=-2^{12}\cdot (4p^3+27q^2)$ is negative, but surprisingly we can explicitly compute the solutions of this equation by the Ferrari’s method.

We introduce a new unknown $u$ so that, using (4),

$$\begin{array}{llll}\hfill {\left(m^2-X+u\right)}^2& ={(m^2-X)}^2+2u(m^2-X)+u^2& \hfill & \\ \hfill & =4m^{2}X+8mY+4X^2+4p+2u{m}^2-2uX+u^2& \hfill & \\ \hfill & =(4X+2u)m^2+8Ym+4X^2-2uX+4p+u^2.& \hfill & \end{array}$$

We search $u$ so that the right member is a perfect square. This will be the case if the discriminant $\mathrm{\Delta }$ relative to $m$ is zero, where

$$\begin{array}{llll}\hfill \mathrm{\Delta }& =8^2{Y}^2-4(4X+2u)(4X^2-2uX+4p+u^2)& \hfill & \\ \hfill & =-64X^3-8u^3+64Y^2-64Xp-32pu& \hfill & \end{array}$$

The change of variable $v=-2u$ gives

$$\begin{array}{llll}\hfill \mathrm{\Delta }& =2^6(v^3+pv+Y^2-X^3-pX)& \hfill & \\ \hfill & =2^6(v^3+pv+q)& \hfill & \\ \hfill & =2^6(v-\alpha )(v-\beta )(v-\gamma ).& \hfill & \end{array}$$

Hence if we take $u=-2v=-2\alpha$, then $\mathrm{\Delta }=0$, and

$$\begin{array}{llll}\hfill {\left(m^2-X-2\alpha \right)}^2& =(4X+2u)m^2+8Ym+4X^2-2uX+4p+u^2& \hfill & \\ \hfill & =4\left[\left(X-\alpha \right)m^2+2Ym+X^2+Xr+{\alpha }^2+p\right]& \hfill & \\ \hfill & & \hfill \end{array}$$

By equalities (1), $\beta +\gamma =-\alpha$ and $\beta \gamma =-\alpha (\beta +\gamma )+p={\alpha }^2+p$, then

$$(X-\beta )(X-\gamma )=X^2+Xr+{\alpha }^2+p,$$

hence

$$\begin{array}{lll}\hfill {\left(m^2-X-2\alpha \right)}^2& =4\left[(X-\alpha )m^2+2Ym+(X-\beta )(X-\gamma )\right].& \hfill \text{(5)}\end{array}$$

We choose complex numbers $A,B,C$ such that

$$\begin{array}{lll}\hfill A^2=X-\alpha ,B^2=X-\beta ,C^2=X-\gamma .& & \hfill \text{(6)}\end{array}$$

Then ${(ABC)}^2=(X-\alpha )(X-\beta )(X-\gamma )=Y^2$, thus $ABC=\pm Y\in R$. We choose $C$ so that $sgn(Y)=sgn(ABC)$. Then $Y=ABC$, and (5) becomes

$$\begin{array}{llll}\hfill {\left(m^2-X-2\alpha \right)}^2& =4[A^2{m}^2+2ABCm+BC]& \hfill & \\ \hfill & =4{(Am+BC)}^2,& \hfill & \end{array}$$

so

$$\begin{array}{lll}\hfill m^2-X-2\alpha =2𝜀(Am+BC),𝜀=\pm 1.& & \hfill \text{(7)}\end{array}$$

Since $-2\alpha =-\alpha +\beta +\gamma =A^2-B^2-C^2$, this gives

$$\begin{array}{lll}\hfill m^2-2𝜀Am+A^2=B^2+C^2+2𝜀BC,& & \hfill \end{array}$$

thus

$$\begin{array}{lll}\hfill {(m-𝜀A)}^2={(B+𝜀C)}^2,& & \hfill \end{array}$$

so

$$\begin{array}{lll}\hfill m& =𝜀A+\eta B+𝜀\eta C,𝜀=\pm 1,\eta =\pm 1.& \hfill \text{(8)}\end{array}$$

Hence the four complex solutions of the quartic equation (4) are

$$\begin{array}{lll}\hfill \begin{array}{cc}{m}_1=A+B+C,\hfill & m_2=A-B-C,\hfill \\ m_3=-A+B-C,\hfill & m_4=-A-B+C,\hfill \end{array}& & \hfill \text{(9)}\end{array}$$

where $A,B,C$ are such that $A^2=X-\alpha ,B^2=X-\beta ,C^2=X-\gamma ,sgn(ABC)=sgn(Y)$.

Explicitly, since $\alpha =r\in ℝ$ and $X\ge r$ for all $P=(X,Y)\in 𝒞$, we take $A=\sqrt{X-r}\in {ℝ}^{+}$. Moreover $B^2=X-\beta =X-s+it$, so

$$\begin{array}{lll}\hfill \begin{array}{cc}\hfill A& =sgn(Y)\sqrt{X-r},\hfill \\ \hfill B& =sgn(Y)\left[\sqrt{\frac{X-s+\sqrt{{(X-s)}^2+t^2}}{2}}+i\sqrt{\frac{-(X-s)+\sqrt{{(X-s)}^2+t^2}}{2}}\right],\hfill \\ \hfill C& =sgn(Y)\left[\sqrt{\frac{X-s+\sqrt{{(X-s)}^2+t^2}}{2}}-i\sqrt{\frac{-(X-s)+\sqrt{{(X-s)}^2+t^2}}{2}}\right].\hfill \end{array}& & \hfill \end{array}$$

Then $sgn(ABC)=sgn(Y)$, and $C=\bar{B}$.

Note that $B\ne C$, otherwise $X-\beta =X-\gamma$, so $\beta =\gamma$, and in this case the discriminant of $p(x)=x^3+a{x}^2+bx+c$ is zero, which is impossible because $𝒞$ is an elliptic curve. Hence $B-C=B-\bar{B}\notin ℝ$. This shows that $m_3$ and $m_4$ are not real. The real solutions are

$$\begin{array}{lll}\hfill m_1& =A+B+C=sgn(Y)\left[\sqrt{X-r}+2\sqrt{\frac{X-s+\sqrt{{(X-s)}^2+t^2}}{2}}\right],& \hfill \text{(10) }\\ \hfill m_2& =A-B-C=sgn(Y)\left[\sqrt{X-r}-2\sqrt{\frac{X-s+\sqrt{{(X-s)}^2+t^2}}{2}}\right].& \hfill \text{(11) }\end{array}$$

(For $Y=0$, and $X=r$, we take $sgn(Y)=1$.)

The corresponding tangent points $(x_1,y_1)$ and $(x_2,y_2)$ are given by the explicit formulas 5.53

$$\begin{array}{llll}\hfill X& =m_1^2-2x_1,Y=-y_1-m_1(X-x_1),& \hfill & \\ \hfill X& =m_2^2-2x_2,Y=-y_2-m_2(X-x_2),& \hfill & \end{array}$$

so

$$\begin{array}{lll}\hfill & x_1=\frac{m_1^2-X}{2},y_1=\frac{1}{2}\left(m_1^3-3X{m}_1-2Y\right),& \hfill \text{(12)}\\ \hfill & x_2=\frac{m_2^2-X}{2},y_2=\frac{1}{2}\left(m_2^3-3X{m}_2-2Y\right).& \hfill \text{(13)}\end{array}$$

Using equations (10) and (12), for $Y>0$, we can define a function $y_1:X\mapsto y_1(X)$ on $[r,+\infty [$ by

$$\begin{array}{llll}\hfill y_1(X)& =\frac{1}{2}\left(m_1{(X)}^3-3X{m}_1(X)-2\sqrt{X^3+pX+q}\right),& \hfill & \\ \hfill \text{where}& & \hfill \\ \hfill m_1(X)& =\sqrt{X-r}+2\sqrt{\frac{X-s+\sqrt{{(X-s)}^2+t^2}}{2}}.& \hfill & \end{array}$$

Then $y_1$ is a continuous function on $[r,+\infty [$. Moreover $y_1(X)\ne 0$ for all $X>r$, and $y_1(0)>0$. By the intermediate value theorem, $y_1(X)>0$ for all $X>r$.

Reasoning similarly with $y_2(X)$, we obtain $y_2(X)<0$ for all $X\ge r$.

By the symmetry of the curve $𝒞$ relative to the $x$-axis, we obtain the same results for $Y<0$.

Therefore the equation $2M=P$, where $P\in 𝒞$ is given and $M=(x,y)$ is the unknown, has two real solutions on $𝒞$, one such as $y<0$ and the other such that $y>0$.

(b)

Construction of the function $\mathbf{\text{P}}$on a dense subset of $ℝ$.

Now we are able to define $\mathbf{\text{P}}$. Put $\mathbf{\text{P}}(0)=P_0=O$, where $O=(0:1:0)$ is the point at infinity of the curve, and more generally $\mathbf{\text{P}}(n)=O$ if $n\in \mathbb{Z}$.

By hypothesis, the polynomial $p(x)$ has only one root $r$. Put $\mathbf{\text{P}}(1∕2)=P_1=(r,0)\in 𝒞$ (then $2\mathbf{\text{P}}(1∕2)=2(r,0)=O$). We build a sequence ${(P_k)}_{k\in {\mathbb{N}}^{\ast }}$ by induction. The points $P_0=O$ and $P_1=(r,0)$ are known. Given $P_k\in 𝒞$, we define $P_{k+1}=(x_{k+1},y_{k+1})$ as the unique point (by part (a)) for which $2P_{k+1}=P_k$ and $y_{k+1}\le 0$. Moreover we define $\mathbf{\text{P}}(1∕2^k)=P_k$. Finally, if $k\in \mathbb{Z}$ is odd, put $\mathbf{\text{P}}(k∕2^s)=k{P}_s$.

Since for all $k\in \mathbb{N}$, $2P_{k+1}=P_k$, we obtain by induction on $u$ that for all $u\in \mathbb{N}$ and for all $s\in \mathbb{N}$,

$$\begin{array}{lll}\hfill P_s& =2^u{P}_{s+u}.& \hfill \text{(14)}\end{array}$$

Suppose that $0\le v\le s$. Then by (6), where $s$ is replaced by $s-v$, $P_{s-v}=2^v{P}_s$, so $P_s=2^{-v}{P}_{s-v}$. This shows that (6) remains true if $-s\le u<0$:

$$\begin{array}{lll}\hfill \forall <!--FUNCTION\; APPLICATION-->s\in {\mathbb{N}}^{\ast },\forall <!--FUNCTION\; APPLICATION-->u\in \mathbb{Z},-s\le u\Rightarrow P_s=2^u{P}_{s+u}.& & \hfill \text{(15)}\end{array}$$

Every $n\in \mathbb{Z}$ has the form $n=k{2}^u$, where $k\in \mathbb{Z}$ is odd and $u\in \mathbb{N}$. For such an integer $n$,

• If $u\le s$,

  $$\begin{array}{llll}\hfill \mathbf{\text{P}}(n∕2^s)& =\mathbf{\text{P}}(k{2}^u∕2^s)& \hfill & \\ \hfill & =\mathbf{\text{P}}(k∕2^{s-u})& \hfill & \\ \hfill & =k{P}_{s-u}& \hfill & \\ \hfill & =k{2}^u{P}_s& \hfill & \\ \hfill & =n{P}_s.& \hfill & \end{array}$$

• If $u>s$,

  $$\begin{array}{llll}\hfill \mathbf{\text{P}}(n∕2^s)& =\mathbf{\text{P}}(k{2}^{u-s})=O,& \hfill & \end{array}$$

  and

  $$\begin{array}{lll}\hfill n{P}_s=k{2}^u{P}_s=k{2}^{u-s}(2^s{P}_s)=O.& & \hfill \end{array}$$

So in both cases $\mathbf{\text{P}}(n∕2^s)=n{P}_s$:

$$\begin{array}{lll}\hfill \forall <!--FUNCTION\; APPLICATION-->n\in \mathbb{Z},\forall <!--FUNCTION\; APPLICATION-->s\in \mathbb{N},\mathbf{\text{P}}(n∕2^s)=n{P}_s.& & \hfill \text{(16)}\end{array}$$

Consider the set

$$A=\{x\in ℝ\mid \exists <!--FUNCTION\; APPLICATION-->n\in \mathbb{Z},\exists <!--FUNCTION\; APPLICATION-->s\in \mathbb{N},x=n∕2^s\}.$$

Then $A$ is a dense subset of $ℝ$, and $\mathbf{\text{P}}$ is defined on $A$. Moreover $A$ is a subring of $ℝ$. We show that $\mathbf{\text{P}}:A\to ℝ$ is a group homomorphism.

Let $x,y$ be some elements of $A$. Then perhaps multiplying one of them by a power of $2$, we my write $x=k∕2^s$, $y=l∕2^s$ with the same denominator $2^s$ (but then $k,l$ are not necessarily odd). Using (8), we obtain

$$\begin{array}{llll}\hfill \mathbf{\text{P}}(x+y)& =\mathbf{\text{P}}((k+l)∕2^s)& \hfill & \\ \hfill & =(k+l)P_s& \hfill & \\ \hfill & =k{P}_s+l{P}_s& \hfill & \\ \hfill & =\mathbf{\text{P}}(k∕2^s)+\mathbf{\text{P}}(l∕2^s)& \hfill & \\ \hfill & =\mathbf{\text{P}}(x)+\mathbf{\text{P}}(y).& \hfill & \end{array}$$

Note that $\mathbf{\text{P}}$ has period $1$ on $A$: if $x=n∕2^s\in A$, then

$$\begin{array}{llllll}\hfill \mathbf{\text{P}}(x+1)& =\mathbf{\text{P}}(k∕2^s+1)& \hfill & & \hfill & \\ \hfill & =\mathbf{\text{P}}((k+2^s)∕2^s& \hfill & & \hfill & \\ \hfill & =(k+2^s)P_s& \hfill (2^s{P}_s=P_0=O)& & \hfill & & \hfill \\ \hfill & =k{P}_s& \hfill & & \hfill & \\ \hfill & =\mathbf{\text{P}}(k∕2^s)& \hfill & & \hfill & \\ \hfill & =\mathbf{\text{P}}(x).& \hfill & & \hfill & \end{array}$$

(c)

Extension by continuity of $\mathbf{\text{P}}$to $ℝ$. We use the following extension theorem (cf. Dieudonné, Treatise of Analysis, t.1, Prop (3.15.6))

Theorem Let $A$be a dense subset of a metric space $E$, and $f$a uniformly continuous mapping of $A$into a complete metric space $E^{\prime }$. Then there exists a continuous mapping $\bar{f}$of $E$into $E^{\prime }$coinciding with $f$in $A$; moreover, $\bar{f}$is uniformly continuous.

Here $\mathbf{\text{P}}:A\to 𝒞$, where $A$ is a dense subset of $ℝ$, with the usual distance $d(x,y)=|y-x|$. What distance are we using on $𝒞$? We take the induced distance on $𝒞$ defined by the metric $\delta$ on the projective plane : we can take for $\delta$ the angle of the lines representing in ${ℝ}^3$ the points of the projective plane (this corresponds to the shortest distance between two points on the Riemann’s sphere $S_2$).

We must prove that $\mathbf{\text{P}}$ is uniformly continuous on $A$ for these chosen distances. We first show that

$$\underset{k\to \infty }{\lim <!--FUNCTION\; APPLICATION-->}{P}_k=O,$$

which is equivalent to

$$\underset{k\to \infty }{\lim <!--FUNCTION\; APPLICATION-->}x(P_k)=+\infty .$$

By part (a), if $P_k=(x_k,y_k)$, then $x_1=r$ and for all $k\ge 1$, $x_{k+1}=\phi (x_k)$, where

$$\begin{array}{lll}\hfill \phi (x)=\frac{{\left[\sqrt{x-r}-2\sqrt{\frac{x-s+\sqrt{{(x-s)}^2+t^2}}{2}}\right]}^2-x}{2}(x\ge r).& & \hfill \text{(17)}\end{array}$$

Since $2P_{k+1}=P_k$, by the explicit formulas (5.53),

$$\begin{array}{lll}\hfill x_k={\left(\frac{3x_{k+1}^2+q}{2y_{k+1}}\right)}^2-2x_{k+1},y_{k+1}=-\sqrt{x_{k+1}^3+p{x}_{k+1}+q}.& & \hfill \end{array}$$

(to be continued ...)