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It turns out this definition can be extended to points of order 2, and also the point O (when we homogenize the functions and work over the projective plane). Moreover, every rational function has as many zeroes as poles counting multiplicities, because of the way we extend the definition to the point at infinity.

I'm interested as to why every rational function has as many zeros as poles. That seems to be caused by "homogenization", so how homogenization works and why do we need it? Why do we need to worry about points of order 2?

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Homogenization refers to the process of going from the "affine" to the "projective" plane, by adding "points at infinity". It's easier to work over the projective plane because there aren't any "missing points"; for example, in the projective plane, every 2 distinct lines intersect at one point (instead of having to make an exception for parallel lines, as in the affine case). In effect, for each class of parallel lines, we add a point at infinity to the plane, and 2 parallel lines will intersect at the point at infinity corresponding to their class; the set of all points at infinity from a line at infinity.

For the study of elliptic curves, adding these missing points is necessary because otherwise, the points of the curve will not form a group (you'll have some missing points, like the point $O$ in the article you linked to).

Here's how the process works. I'll just show how it works in the plane; the same thing works in any number of dimensions. In the usual ("affine") plane, a point is given by coordinates $(x,y)$. In the projective planes, the coordinates will be written $(x:y:z)$. However, unlike the affine case, not all sets of coordinates will be distinct points; in the projective plane,

  1. (0:0:0) is not a point.
  2. If 2 coordinates differ by a scalar multiple, we consider them the same point: $(x:y:z) \sim (ax:ay:az)$. Here $a \ne 0$, and $x,y,z$ cannot all be 0, because of condition 1.

Just like in the affine case, we can write equations of curves using the projective coordinates $(x:y:z)$. However, we have to be careful, because of the equivalence condition imposed by condition 2 above. We cannot write a projective equation like $x+y+z=1$, because if we scale everything by a scalar $a$, we would get $ax+ay+az=1$, which is not equivalent to $x+y+z=1$. Indeed, condition 2 limits us to equations of the form $f(x,y,z)=0$ where $f$ is a homogeneous polynomial, meaning all terms of the polynomial have the same degree. For example, $y^2z - x^3 - 2xz^2 - z^3 = 0$ is a legitimate equation for a curve in the projective plane, because if we scale each coordinate by a scalar $a$, all the terms are multiplied by $a^3$, which can then be cancelled out (because $a \ne 0$ and the right-hand side is 0!). You can "homogenize" an affine equation in coordinates $x,y$ by multiplying each term by an appropriate power of $z$. That's what I did to get the equation above; starting with the affine elliptic curve equation $y^2 - x^3 - 2x - 1 = 0$, I homogenized each term to get the projective equation.

Because of condition 2, whenever we have a point $(x:y:z)$ and $z \ne 0$, we can normalize the $z$ coordinate to 1: $(x:y:z) \sim (x/z, y/z, 1)$. So among the points with non-zero $z$ coordinate, we effectively have only 2 coordinates to worry about, and we are back in the affine case. So the affine plane is contained in the projective plane: $(x,y) \mapsto (x:y:1)$. But the projective plane also contains an additional set of points with $z$-coordinate 0. These are the points at infinity.

In the affine plane, the lines $x+y=1$ and $x+y=2$ are parallel. Homogenizing, we get $x+y=z$ and $x+y=2z$ in the projective plane, where they intersect at a point at infinity (1:-1:0). (You can check that any other solution to this pair of equations is equivalent, via condition 2, to (1:-1:0), so there is only one intersection point in the projective plane.) The point (1:-1:0) is the point at infinity corresponding to (affine) lines of slope 1. We added one point an infinity for each possible slope (including $\infty$ for vertical lines); (1:m:0) for lines of slope $m$, and (0:1:0) for vertical lines.

In the elliptic curve case, homogenizing adds the point (0:1:0) to the curve, which is the point $O$ in the article linked in your original post; this point is usually taken to be the identity of the group. As noted above, (0:1:0) is the point at infinity corresponding to vertical lines. If you take an affine elliptic curve such as $y^2 - x^3 - 2x - 1 = 0$ and try to add the 2 points $(x,y)$ and $(x,-y)$, when you draw the line through them, they don't intersect the curve in a third point. But in the projective plane, they do; $(x,y)$ and $(x,-y)$ lie on a vertical line and therefore the line through them contains the point (0:1:0).

The situation with rational functions is similar. I'll switch down to 1 dimension since that illustrates the point more clearly. In the affine line, with a single coordinate $x$, I can define the rational function $1/x$. This has a pole at $x=0$ but no zeroes. However, on the projective line (with projective coordinates $(x:y)$), I homogenize to the function $y/x$. Now there is a pole (0:1) as before but there is also the zero (1:0). There are the same number of zeroes as poles. This is true for every rational function (not identically zero) on the projective line. It is also true for elliptic curves and indeed for every projective curve.

Note: This answer may give the impression that the $z$-coordinate is somehow "special" in the projective plane. This isn't really the case; it's an artifact coming from the way I chose to embed the affine plane into the projective plane, via $(x,y) \mapsto (x:y:1)$. This is convenient for relating the projective plane back to the familiar case of the affine plane, but it's just a choice, sort of like choosing a basis for a vector space; I could also choose either of the other coordinates to be the special one, or even something crazy like $(x,y) \mapsto (x+y+1:2x+3y+4:4x+5y+7)$ (not that one would actually want do something like that, which would be like deliberately choosing a weird basis for a vector space).

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let $f(x)$ be a polynomial. $f$ has a pole at infinity if $f(1/x)$ has a pole at zero (is that correct?). Now let $f(x,y)=p/q$ be a rational function and let $F$ be its homogenization. How do we define a pole for $F$? –  ted.k Sep 17 '11 at 22:17
    
A pole for $F$ is a point for which the homogenized denominator is 0. (We assume that $F$ is expressed as the ratio of relatively prime homogeneous polynomials.) Note that here there are 2 variables involved, so the poles will actually form a curve and not just be isolated points. But if you are considering $F$ as a rational function on the elliptic curve rather than the whole plane, then you have to intersect these "pole curves" with the elliptic curve, which will again give you isolated points on the elliptic curve. (Except when the "pole curve" is the entire elliptic curve.) –  Ted Sep 18 '11 at 6:57
    
so let's say $F=(x-1)/(x+1)$ and there is no point on our curve with $(-1,y)$. What now? I don't seem to get the part as to why the denominator is guaranteed to have as many zeros as numerator, on the curve :(. Is it because of Bezout's theorem about intersections? –  ted.k Sep 18 '11 at 7:32
    
Even a simpler example: let $F$ be a polynomial. So we have a homogenized polynomial of three variables (X,Y,Z) with no denominator. –  ted.k Sep 18 '11 at 9:26
    
Say $f = x^3$; then the homogenized version is $F = x^3/z^3$. In general, the homogenization of $f(x,y)$ is $f(x/z, y/z)$. (Remember that if $z \ne 0$, the projective point $(x:y:z)$ corresponds to the affine point $(x/z,y/z)$.) Regarding your other comment, there can't be "no point" on the curve with $(-1,y)$, because of Bezout's theorem; if the field is algebraically closed and you're working over the projective plane, a curve of degree $m$ and degree $n$ with no common components always have exactly $mn$ intersections counting multiplicity. –  Ted Sep 18 '11 at 17:10
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Given a rational function $f$ on an elliptic curve (or a smooth projective curve) $E$, we have that $$\sum_{x\in E} \textrm{ord}_x(f) =0. $$ If $\textrm{ord}_x(f) >0$, we say that $f$ has a zero of order $\textrm{ord}_x(f)$ at $x$. If $\textrm{ord}_x(f)<0$, we say that $f$ has a pole of order $-\textrm{ord}_x(f)$ at $x$. The statement I just gave is usually phrased as "the degree of a principal divisor is zero".

Here $\textrm{ord}_x$ is the discrete valuation on (the fraction field of) the local ring of $E$ at $x$. If you want to compute it for a rational function $f$ which is regular on an open affine $U$ containing $x$ you can choose an element $\pi \in \mathcal{O}(U)$ which generates the maximal ideal of $\mathcal{O}_{X,x}$ and write $f= u \pi^n$ on $U$ with $u$ a unit in the local ring. Then $\textrm{ord}_x(f) = n$. In general, to compute the order of $f$ at $x$ one writes $f$ as a quotient of two regular functions in an open affine neighborhood of $x$.

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Sorry for the possible irritation, but is there a proof for the fact that the degree of principal divisors is o? Or does this follow from the product formula on the residue field of the local ring? I am quite surprised to find out that this result (the numbers of poles and of zeros coincide), which appears to be pertinent to the analysis, is in fact a statement about the local field, which topic I have incidentally acquired much more than I have for the analysis. In sum, I am inspired by this great answer, albeit dearth of a formal proof, which I think is easy, is it not? –  awllower Dec 14 '11 at 13:55
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