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.