I'm studying basics of elliptic curves. I'm reading An Elementary Introduction to Elliptic Curves by Leonard Charlap and David Robbins. It is stated there that the divisor of a line (i.e. a polynomial of the form $ax + by + c$) can have only few forms, among them is $3\langle P \rangle – 3\langle \mathcal{O}\rangle$. I tried to find an example of a curve and a line on it that has such divisor, but to no avail. Can anyone provide an example? If it helps, they suggest that $P$ is an inflection point.
[Math] Elliptic curves, inflection points and divisors
elliptic-curves
Related Solutions
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".
Best Answer
Let the base field be $F_2$ (hopefully you're fine with a finite base field). Let the curve be $y^2+y=x^3$ and the line $y=0$. The function $y$ has a pole of order 3 at the point of infinity ${\mathcal O}$ and a triple zero at origin ${\mathcal P}=(0,0)$, so the divisor of $y$ is $3{\mathcal P}-3{\mathcal O}$ as prescribed.
Edit: D'oh. The OP asked for examples in other characteristics. I'm apparently at a my dullest. Doesn't the same example work in any characteristic? (Except at char 3, because then the curve has a singular point). See a figure of the real points below.
The origin looks like an inflection point to me :-)