Timeline for Intuition for the last step in Serre's proof of the three-squares theorem
Current License: CC BY-SA 4.0
11 events
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| Jul 7, 2023 at 7:27 | history | edited | KConrad | CC BY-SA 4.0 |
added 62 characters in body
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| Aug 24, 2018 at 14:33 | comment | added | David E Speyer | Very late comment, but I think the computation in $\mathbb{Z}^4$ still works if you round to the nearest point in $\mathbb{Z}^4 \cup \left( \mathbb{Z}^4 + (1/2,1/2,1/2,1/2) \right)$. | |
| Jan 16, 2010 at 6:26 | comment | added | KConrad | It's stronger only because of the non-archimedean behavior of the degree on F(T), and that removes the surprise for me. What happens is that in Q, a sum of k numbers that are each at most 1/4 is less than 1 only for k = 1, 2, and 3, but in F(T) any finite sum of rational functions with degree below some bound also has degree below that bound. | |
| Jan 16, 2010 at 5:40 | comment | added | Bjorn Poonen | OK, thank you. I hadn't thought about the function field case before - it's interesting that you get a stronger result in that case! | |
| Jan 16, 2010 at 4:11 | comment | added | KConrad | When I gave this proof in a lecture (to students at the Ross Program at Ohio State in 2003), Markus Rost was in the audience. He brought to my attention that instead of speaking about the second point of intersection of a line with a circle, you could speak about reflections (for circles, not ellipses!). This led me to write up the details of the argument using the language of reflections, and although it only works up to a sum of 3 squares in Z, it goes through for n squares in F[T]. Look here: math.uconn.edu/~kconrad/blurbs/linmultialg/sumsquareQF(T).pdf Is that helpful? | |
| Jan 16, 2010 at 3:54 | comment | added | KConrad | I pointed out the example of x^2 + 82y^2 = 2 in my answer to illustrate that intuition has to be supplemented by an actual argument, since it doesn't seem like intuition can distinguish between x^2 + 82y^2 = 2 and x^2 + y^2 = a. | |
| Jan 16, 2010 at 3:51 | comment | added | KConrad | The point I was trying to make was that the denominators drop "because" you're connecting (by a line) a point with denominators greater than 1 to a point whose denominators all equal 1, so the new rational point that comes out of the process has denominator in between the two. I think the fact that the coordinates of the nearby integral point are all 1 is the intuition for why this process makes the denominators go down from one rational solution to the next. The question asked for intuition about the process, not a conceptual proof of the method. I don't have a conceptual proof. :( | |
| Jan 16, 2010 at 2:00 | comment | added | Bjorn Poonen | Hi KConrad; yes, the lcm of the denominators gets closer to 1 at each step (as Qiaochu said already in the question statement). The unanswered question is whether there is a conceptual proof of this that is any clearer than the proofs given so far. For example, can one make it conceptually clear why the relevant hypothesis on the quadratic form is that the absolute value of its value at the "error vector" in the lattice point approximation should be between 0 and 1? | |
| Jan 16, 2010 at 0:11 | history | edited | KConrad | CC BY-SA 2.5 |
fixed grammar
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| Jan 15, 2010 at 12:23 | vote | accept | Qiaochu Yuan | ||
| Jan 15, 2010 at 4:52 | history | answered | KConrad | CC BY-SA 2.5 |