# Elliptic Curves and Points at Infinity

My undergraduate number theory class decided to dip into a bit of algebraic geometry to finish up the semester. I'm having trouble understanding this bit of information that the instructor presented in his notes.

Here it is in paraphrase (assume we are over an abstract field k)

We take a polynomial in k, $f =Y^2 - X^3 -aX -b$ and homogenize the polynomial to $F = Y^2Z - X^3 -aXZ^2 - bZ^3$. Note that the points at infinity of V(F) consist of triples $[\alpha : \beta : 0]$ s.t $-\alpha^3 = 0$, hence the only point at infinity is $[0 : 1 :0]$

The part I'm confused about is in italics. He introduces the terms "points at infinity" without defining it. After some google time, I understand what a point at infinity means in the context of a projective space/projective line but am having trouble understanding how the professor came to his conclusion about the point at infinity in this particular example

Here is my question. In general, are all points in the locus of vanishing points for a homogeneous polynomials considered points at infinity? If not, is there a general procedure for calculating these point if we are given an arbitrary polynomial?

More abstractly, How do I understand that a finite point in the projective space is a "point at infinity" for this polynomial.

-
@xdfm Yes, they are. You can imagine my surprise when I missed thanksgiving week and we jumped infinite fractions to discussing elliptic curves. – crasic Dec 10 '10 at 5:48
I would recommend you to read the first few sections of the appendix in Rational Points on Elliptic Curves and the first chapter of the book. It explains all this matters quite nicely and at a very elementary level. – Adrián Barquero Dec 10 '10 at 6:27
@Adrián this book is at the perfect level. Thank you for the link. – crasic Dec 10 '10 at 7:27
I'm glad you like it. It is one of my favorite books. I hope you enjoy it as much as I did. – Adrián Barquero Dec 10 '10 at 21:16

Here's another way to think about the "line at infinity" and the "points at infinity"...

Think of the usual $XY$-plane as sitting inside of $3$-space, but instead of it sitting in its usual place, $\{(x,y,0) : x,y\in\mathbb{R}\}$, shift it up by $1$ so that it sits as the $z=1$ plane.

Now, you are sitting at the origin with a very powerful laser pointer. Whenever you want a point on the $XY$-plane, you shine your laser pointer at that point. So, if you want the point $(x,y)$, you are actually pointing your laser pointer at the point $(x,y,1)$; since you are sitting at the origin, the laser beam describes a (half)-line, joining $(0,0,0)$ to $(x,y,1)$.

Now, for example, look at the point $(x,0,1)$, and imagine $x$ getting larger. The angle your laser pointer makes with the $z=0$ plane gets smaller and smaller, until "as $x$ goes to infinity", your laser pointer is just pointing along the line $x$ axis (at the point $(1,0,0)$), and the same thing happens if you let $x$ go to $-\infty$. More generally, if you start pointing to points that are further and further away from the "origin" in your plane (away from $(0,0,1)$), the laser beam's angle with $Z=0$ gets smaller and smaller, until, "at the limit" as $||(x,y)||\to\infty$, you end up with the laser beam pointing along the $z=0$ plane in some direction. We can represent the direction with the slope of the line, so that we are pointing at $(1,m,0)$ for some $m$ (or perhaps to $(-1,-m,0)$, but that's the same direction), or perhaps to the point $(0,1,0)$. So we "add" these "points at infinity" (so called because we get them by letting the point we are shining the laser beam on "go to infinity"), one for each direction away from the "origin": $(1,m,0)$ for arbitrary $m$ for nonvertical lines, and $(0,1,0)$ corresponding to the direction of $x=0$, $y\to\pm\infty$.

So: the "usual", affine points, are the ones in the $z=1$ plane, and they correspond to laser beams coming from the origin; they are each of the form $(x,y,1)$ for some $x,y$ in $\mathbb{R}$. In addition, for each "direction" we want to include that limiting laser beam which does not intersect the plane $z=1$; those correspond to points $(1,m,0)$, or the point $(0,1,0)$ when you do it with the line $x=0$. So we get one point for every real $m$, $(1,m,0)$, and another for $(0,1,0)$. You are adding one point for every direction of lines through the origin; these points are the "points at infinity", and together they make the "line at infinity".

Now, put your elliptic curve/polynomial $F=Y^2 - X^3 - aX-b$, and draw the points that correspond to it on the $z=1$ plane; that's the "affine piece" of the curve. But do you also get any of those "points at infinity"?

Well, even though we are thinking of the points as being on the $XY$-plane, they "really" are in the $Z=1$ plane; so our equation actually has a "hidden" $Z$ that we lost sight of when we evaluated at $Z=1$. We use the homogenization $f = Y^2Z - X^3 - aXZ^2 - bZ^3$ to find it. Why that? Well, for any fixed point $(x,y,1)$ in our "$XY$-plane", the laser pointer points to all points of the form $(\alpha x,\alpha y,\alpha)$. If we were to shift up our copy of the plane from $Z=1$ to $Z=\alpha$, we'll want to scale everything so that it still corresponds to what I'm tracing from the origin; this requires that every monomial have the same total degree, which is why we put in factors of $Z$ to complete them to degree $3$, the smallest we can (making it bigger would give you the point $(0,0,0)$ as a solution, and we do need to stay away from that because we cannot point the laser pointer in our eye).

Once we do that, we find the "directions" that also correspond to our curve by setting $Z=0$ and solving, to find those points $(1,m,0)$ and $(0,1,0)$ that may also lie in our curve. But the only one that works is $(0,1,0)$, which is why the elliptic curve $F$ only has one "point at infinity".

-
A concise anwer that I can understand: very valuable. "we cannot point the laser pointer in our eye": priceless! – David Schmitt Dec 10 '10 at 13:08
I'm also trying my hand at elliptic curves recently. This is a great post! – yunone Dec 10 '10 at 19:39

The term "point at infinity" is not actually a well-defined term from the point of view of your projective model. You should think of it this way:

Suppose you are given a homogeneous polynomial and you are interested in its zero set as a subset of projective space. The projective space is made up of several so-called affine pieces, and accordingly your zero set is made up of several affine pieces. Here is how it works: let's take the concrete polynomial $F=Y^2Z - X^3-aXZ^2 - bZ^3$. Since you can take any point $(x:y:z)$ satisfying this polynomial and scale the coordinates by any non-zero scalar, $\alpha$ say, and the result still represents the same point, you may scale the point by $\alpha=z^{-1}$ and write it as $(x/z,y/z,1)$. But now, you have to be careful: this only works for points, at which $z\neq 0$. For those points, you are setting $z=1$, so the resulting affine model is given by your polynomial $f$ (just set $Z=1$ in $F$).

If on the other hand $z$ happened to be 0, then you have attempted to divide by 0, so in this sense, you obtained a "point at infinity". Now, what is this point at infinity? Set $Z=0$ in $F$ and see what points satisfy the resulting polynomial. For the polynomial to be 0, you also need $-X^3=0$, so $X=0$. Now, since we are in projective space, $Y$ mustn't be 0 (recall that (0:0:0) is not a point in projective space), so you can scale the coordinates so that $Y=1$. That's your "point at infinity" for this particular affine piece: $(X:Y:Z)=(0:1:0)$.

But the procedure was not canonical. Instead, you might have chosen to look at the affine piece $Y\neq 0$, say. Then, you would have written all points in this affine model as $(x/y:1:z/y)$ and the dehomogenised polynomial would have been different. Also, the "points at infinity" would have been different ones, namely all those projective points, for which $Y=0$.

I hope this makes sense. Otherwise, feel free to ask for clarifications.

-
If I'm understanding you correctly, the "infinity" comes when we try to map this point back into k^n, over which the polynomial is defined? In other words, if we homogenize a polynomial (and Z is the added dimension), then all points in the vanishing locus with Z = 0 are points at infinity? In the end, isn't our goal to analyse the polynomial itself (and not the homogenized version)? So it makes sense to "normalize" with respect to Z. – crasic Dec 10 '10 at 5:45
@crasic: whether it's our goal to analyze the polynomial itself depends on your point of view. When you look at the elliptic curve as a group, you need to take into account the point at infinity because it forms the identity for the group operation. It's just that looking at a "finite" slice is convenient for geometric intuition. – Qiaochu Yuan Dec 10 '10 at 5:51
To add to what Qiaochu said: the elliptic curve is more accurately thought of as being given by the projective polynomial, rather than the affine one. So, you should think of $F$ as the starting point, and not of $f$. Now, we usually find affine space easier to work with than projective space, so we consider an affine piece and the points on the curve in that affine piece. Which affine piece you want to consider depends in the concrete situation. In this particular case, it is easy to see that if you consider the piece $Z\neq 0$, then you only miss one point, while if instead you considered... – Alex B. Dec 10 '10 at 5:55
...$Y\neq 0$, you would have to undertake a careful analysis of what you have left out. So you naturally decide to do the former. Because this is so common, you will often see people talking about an elliptic curve, when they mean this particular affine piece of it. But they always remember in the back of their mind, that they have left out one projective point - the "point at infinity". – Alex B. Dec 10 '10 at 5:57