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In article Adams, W. W., & Razar, M. J. (1980). Multiples of points on elliptic curves and continued fractions. Proc. London Math. Soc, 41, 481-498. is said on page 482 that elliptic curve

$$E_a: w^2=v^3+Av+B$$

has one point $O$ at infinity.

I understand it in such way: there is projective curve

$$E_p: uw^2=v^3+Au^2v+Bu^3$$

and points that in $E_p - E_a$ (as sets) are points at infinity.

In our case there is only one point, which third coordinate equal to $0$: $(0:1:0)$ (we set $u=0$ in $E_p$ polynomial, it automatically forces that $v=0$).

Then authors let $P = (a,b) \ne O$ and

construct a plane quatric model for E, with respect to which $O$ and $P$ are the two points (valuations) at infinity. Let

$x = \frac{w+b}{v-a}, y = 2v+a - (\frac{w+b}{v-a})^2$

after "some algebra" they have

$$y^2=x^4-6ax^2-8bx+c$$

where $c=-4A-3a^2$.

Later they assert that there are two points at infinity $P$ and $O$.

If we look only at last equation we can get only one infinity point -- again (0:1:0). The second should arrive somehow from the map, but I don't understand how.

So the question is: how to explain existence of two points on infinity after change of variables? And how to calculate points on infinity in general case? And how to interpret map between projective varieties in terms of map of affine varieties (like in this case)?

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  • $\begingroup$ This is not an MO question. The function field of the quartic has two embeddings in $k((1/x))$ corresponding to $y = \pm x^2 + \cdots$, so two places at infinity. $\endgroup$
    – Felipe Voloch
    Commented Dec 4, 2014 at 17:11

3 Answers 3

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If you just take the projective closure in ${\mathbb P}^2$, then your curve will have only one point at infinity, but this point is singular. What you really want to consider is the smooth projective model of the curve, which you obtain by resolving the singularity at infinity. This leads to two distinct points.

One concrete way of doing this is to consider the weighted projective plane ${\mathbb P}^2_{1,2,1}$ and coordinates $x,y,z$; then the homogeneous equation is $$y^2 = x^4 - 6 a x^2 z^2 - 8 b x z^3 + c z^4$$ and the points at infinity on this (smooth) model are $(1 : \pm 1 ; 0)$.

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  • $\begingroup$ Thanks! Where is it possible to read about weighted projective plane? It is quite new notion for me. $\endgroup$
    – Maxim
    Commented Dec 4, 2014 at 16:16
  • $\begingroup$ One more question: there was smooth curve and then we map it to curve with singularity at infinity. So we created singularity in some case. What is the general approach to such situations? $\endgroup$
    – Maxim
    Commented Dec 4, 2014 at 16:20
  • $\begingroup$ To get the points of ${\mathbb P}^2_{1,2,1}$ you identify coordinates $(x,y,z)$ and $(\lambda x, \lambda^2 y, \lambda z)$ (with $\lambda \neq 0$ as usual). The degree of a polynomial (w.r.t. the weights) is then defined by using $\deg(x) = 1$, $\deg(y) = 2$, $\deg(z) = 1$, so that the equation in my answer is homogeneous of degree 4 and therefore defines a subvariety of ${\mathbb P}^2_{1,2,1}$. The advantage of using these weighted spaces is that one can get smooth models in lower-dimensional projective spaces. A disadvantage is that weighted projective spaces can have ... $\endgroup$
    – Michael Stoll
    Commented Dec 4, 2014 at 17:35
  • $\begingroup$ ... singular points (which are not on the curve above, however). But, as Joe Silverman points out, it is possible to avoid the weighted projective plane here and work with a collection of affine patches instead. $\endgroup$
    – Michael Stoll
    Commented Dec 4, 2014 at 17:37
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    $\begingroup$ As to your second comment, the singuarity shows up because you do not take the "right" projective closure. If you have a smooth affine curve, then there is always a smooth projective curve (sitting in ${\mathbb P}^n$ for some $n$) that is birational to it; this smooth projective model is unique up to isomprphism. The "points at infinity" then are the points one has to add to the affine curve to obtain the smooth projective model. $\endgroup$
    – Michael Stoll
    Commented Dec 4, 2014 at 17:41
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Possibly it's easier to take two affine curves and glue them together, rather than taking weighted projective space (which admittedly is slicker). So take $$ C_1 : y^2 = x^4 + ax^3 + bx^2 + cx + d $$ and $$ C_2 : v^2 = 1 + au + bu^2 + cu^3 + du^4 $$ and glue them together (along appropriate subsets) via $$ x = 1/u\quad\text{and}\quad y=v/u^2. $$ The points at infinity on the smooth projective closure of $C_1$ are the points on $C_2$ with $u=0$, which are the two points $(0,1)$ and $(0,-1)$.

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Elliptic curves double cover the Riemann sphere with 4 branch points. When you look at an affine model of the elliptic curve, naturally the points at infinity are missing. If the point at infinity is not one of the branch points, it will correspond to two points on the elliptic curve that are missing. That's a way of seeing it without writing down any formulas.

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