Here’s a basic example that comes up if you work with elliptic curves: Let be a prime which is . Let be the elliptic curve over a field of characteristic . Then has an endomorphism . It turns out that, in the group law on , we have . That is to say, plus copies of is trivial.
I remember when I learned this trying to check it by hand, and being astonished at how out of reach the computation was. There are nice proofs using higher theory, but shouldn’t you just be able to write down an equation which had a pole at and vanished to order at ?
There is a nice way to check the prime by hand. I’ll use for equivalence in the group law of . Remember that the group law on has and has whenever , and are collinear.
We first show that
Proof of : We want to show that , and add up to zero in the group law of . In other words, we want to show that these points are collinear. We just check:
as desired. .
Use of : Let be a point on . Applying twice, we get
Now, the horizontal line crosses at three points: , and . (Of course, , since we are in characteristic three.) So and we have
as desired. .
I was reminded of this last year when Jared Weinstein visited Michigan and told me a stronger statement: In the Jacobian of , we have , where is once again the automorphism .
Let me first note why this is related to the discussion of the elliptic curve above. (Please don’t run away just because that sentence contained the word Jacobian! It’s really a very concrete thing. I’ll explain more below.) Letting be the curve , and letting be , we have a map sending , and this map commutes with . I’m going to gloss over why checking on will also check it on , because I want to get on to playing with the curve , but it does.
So, after talking to Jared, I was really curious why acted so nicely on the Jacobian of . There are some nice conceptual proofs but, again, I wanted to actually see it. Now I do.
Let be any odd prime. Let be the curve , over a field of characteristic . We’ll be working in the Jacobian of . This is a group , generated by the points of , and subject to the relation that, if there is a polynomial vanishing precisely at the points , , …, , then . If you’ve seen this theory laid out for projective curves, then you use rational functions rather than polynomials, and you have to keep track of poles as well as zeroes. Because I’m using the affine curve , I get to just work with polynomials and their zeroes.
When we run this construction for a curve of the form , then the elements of the group are just the points on the curve, plus the additive identity. Let’s see why: , as these are the two points on the line , so we don’t need formal inverses. And for any two points and with , the line through these points will meet the curve once more, say at , so we have . We can repeatedly use this trick to reduce sums of many points to sums of fewer points.
I am now asking you to consider , which (except when ) is not of the form
. This means that not every element of will be equivalent to a single point of the curve . Other than that, the theory really isn’t harder.
The curve has an endomorphism , just as before. And, just as before, what we are going to be showing is that as endomorphisms of the group . In particular, when , we have .
The key identity
We’re going to figure out how to generalize equation for all primes. The equation we want is
Here is the quadratic residue symbol: it equals if is a quadratic residue modulo ; it equals if is a non-QR and if .
Just as in the case, we’ll want to rewrite this to have fewer minus signs. For any on , the vertical line meets at two points: and . So . So we can rewrite our desired equation as
So we need to find a polynomial which passes through the points . Set
Since , the polynomial will pass through all of the points . Where else does meet ?
Well, if then . Plugging this into the equation for , we have
This is a degree polynomial in . We already know of the roots — they are at for a nonzero element of .
Equation has leading term , and constant term . So we can conclude that the product of the roots of is . We have . So the last root of is at . A little more thought shows that the intersection of with is at , not .
So we have found a polynomial which vanishes precisely at the points and , and we have proved equation .
Why we’ve won
Let be the automorphism of the curve . Clearly, . A little less obviously, I claim that in the endomorphism ring of the group . In other words, I am claiming that, for any on , we have
This is because the points are precisely the intersections of with the horizontal line .
So is a module for the ring . This is better known as the ring of cyclotomic integers. And identity tells us that
in the ring .
The right hand side of the above equation is a very famous element of : The Gauss sum. And it’s most famous property is that — exactly what we wanted to show.
Let’s review why the Gauss sum has the desired square.
Group together terms with the power of to get
If , then takes on every value in once, except that it misses . If , then takes on the value over and over, times in total. Using , our sum is
What I want to emphasize is that every equality of this proof corresponds to writing down a rational function on with the corresponding poles and zeroes. For example, the second to last equality above replaced by . There is a corresponding rational function which has zeroes at and has a pole at : Namely, . It would be painful, but completely doable, to actually use this proof to write down a rational function on with a zero at , and a -fold pole at . In fact, I cheated a bit in the beginning of this post. I didn’t just cleverly guess the formula ; I instead wrote down and plugged in .
Descending the resulting rational function to the curve would probably leave an unmotivated mess. I understand how to do it: You need to look at the rational function of and rewrite it in the variables ; Galois theory guarantees that you will succeed. But I suspect the result would be unenlightening.
So, finally, I understand why without going through the theory of supersingular curves, or the Weil bounds, or anything deeper than Gauss sums. Hope you enjoyed!