Topological methods for some arithmetic questions

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1 Topological methods for some arithmetic questions GilYoung Cheong University of Michigan Ongoing project: UM Math Club talk September 21, 2016 GilYoung Cheong (University of Michigan) September 21, / 41

2 Overview 1 Introduction Weil s three columns Review of terminology An arithmetic question 2 Zieve s argument 3 Ellenberg s argument What are the Betti numbers of a space? 4 Generalizations Non-vanishing conditions = Puntures Computation of the cohomological information 5 Further question GilYoung Cheong (University of Michigan) September 21, / 41

3 Weil s three columns In Ed Frenkel s book Love and Math, the author quotes André Weil s letter to his sister where he explains mysterious connections among the following three columns. Number Theory Curves over Finite Fields Riemann Surfaces GilYoung Cheong (University of Michigan) September 21, / 41

Weil s three columns My work consits in deciphering a trilingual text; of each of the three columns I have only disparate fragments; I have some ideas about each of the three languages: but I know as

4 Weil s three columns My work consits in deciphering a trilingual text; of each of the three columns I have only disparate fragments; I have some ideas about each of the three languages: but I know as well there are great differences in meaning from one column to another, for which nothing has prepared me in advance. In the several years I have worked at it, I have found little pieces of the dictionary. GilYoung Cheong (University of Michigan) September 21, / 41

5 Weil s three columns What we think about today is roughly a connection between the last two columns. In this talk, it is okay to think that topologial = geometric. GilYoung Cheong (University of Michigan) September 21, / 41

6 Review of terminology Let p be a prime number. A finite field of p elements is simply the set F p := {0, 1,, p 1}. Addition / Multiplication. In F 5 = {0, 1, 2, 3, 4}, we require 5 = = 7 = 2; 4 4 = 16 = 1. GilYoung Cheong (University of Michigan) September 21, / 41

7 Review of terminology In other words, we see F p is the set of integers Z with the relation p = 0. Why prime p? To have DIVISION! If we require 6 = 0, then What s wrong here? 2 3 = 6 = 0. GilYoung Cheong (University of Michigan) September 21, / 41

8 Review of terminology In general, a set F where you can add, subtract, multiply, and divide is called a field. For example, 1 the set R of real numbers; 2 the set C = R 2 of complex numbers; 3 the set F p of integers modulo a prime number p. GilYoung Cheong (University of Michigan) September 21, / 41

9 Review of terminology NOT fields. 1 the set Z of integers (e.g., 1/2 / Z); 2 the set F [x] of polynomials in the variable x whose coefficients are in a field F (e.g., 1/x / F [x]), but still they have... GilYoung Cheong (University of Michigan) September 21, / 41

10 Review of terminology Prime factorization! Given f (x) F [x], we can uniquely factor f (x) = q 1 (x) n1 q k (x) n k where q i (x) F [x] are primes. If n 1 = n 2 = = n k = 1, we call f (x) a square-free polynomial. GilYoung Cheong (University of Michigan) September 21, / 41

11 Review of terminology For example, you can check that the monic (the leading coeffcient = 1) polynomial x 2 + x + 1 R[x] is square-free, because it is an irreducible itself. (Why?) BUT x 2 + x + 1 = x 2 2x + 1 = (x 1) 2, so the polynomial is NOT square-free in F 3 [x]. GilYoung Cheong (University of Michigan) September 21, / 41

12 Review of terminology Exercise. Can you tell whether the (monic) polynomial is square-free in F 2 [x]? x 2 + x + 1 GilYoung Cheong (University of Michigan) September 21, / 41

13 Review of terminology Exercise. How many monic polynomials of degree 2 in F 2 [x]? Exercise. Given a prime p and an integer d 0, how many monic polynomials of degree d in F p [x]? Exercise. Given a prime p, how many monic square-free polynomials of degree 2 in F p [x]? GilYoung Cheong (University of Michigan) September 21, / 41

14 An arithmetic question Question Given a finite field F p = {0, 1,, p 1} with a prime p, what is the number of monic square-free polynomials with degree d 0 in F p [x]? GilYoung Cheong (University of Michigan) September 21, / 41

15 An arithmetic question: counting degree d monic square-free polynomials in F p [x] For d = 0, the answer is 1; For d = 1, the answer is p. Do you see why? For d = 2, we saw p 2 p; For d = 3, the counting becomes a bit more complicated. GilYoung Cheong (University of Michigan) September 21, / 41

16 An arithmetic question: counting degree d monic square-free polynomials in F p [x] Spoiler: the answer is p d p d 1 for d 2. TWO distinct arguments! 1 U Michigan: Michael Zieve s argument; 2 U Wisconsin: Jordan Ellenberg s argument. FYI. Zieve and Ellenberg have been friends for many years! GilYoung Cheong (University of Michigan) September 21, / 41

17 Zieve s argument We first consider the case d = 3. Consider the map monic, NOT sq-free F p [x] deg=3 F p [x] monic deg=1 = {x + a : a F p} given as follows. Since F p [x] enjoys the unique factorization, any monic polynomial f (x) that is NOT square-free in it can be uniquely written as f (x) = (x + a)(x + c) 2. GilYoung Cheong (University of Michigan) September 21, / 41

18 Zieve s argument Our map is then given by (x + a)(x + c) 2 x + a, where g(x) is monic square-free of degree 2. Moreover, any polynomials of the following form map to x + a: which has p elements. {(x + a)(x + c) 2 : c F p }, GilYoung Cheong (University of Michigan) September 21, / 41

19 Zieve s argument Thus, the map is a p-to-1 surjective map, which implies and hence #{monic NOT square-free polynomials of degree 3} = p #{monic polynomials of degree 1} = p 2, #{monic square-free polynomials of degree 3} = p 3 p 2. Q.E.D. GilYoung Cheong (University of Michigan) September 21, / 41

20 Exercise. When you go home, generalize Zieve s argument to conclude that for any d 2, we have monic, sq-free #F p [x] deg=d = p d p d 1. We now consider Ellenberg s topological argument. GilYoung Cheong (University of Michigan) September 21, / 41

21 Ellenberg s argument We consider the case d = 2. First, notice that we have the following bijection given by {monic polynomials of degree 2} A 2 (F p ) := F 2 p x 2 + t 1 x + t 2 (t 1, t 2 ). GilYoung Cheong (University of Michigan) September 21, / 41

22 Ellenberg s argument The polynomial x 2 + t 1 x + t 2 is square-free if and only if its discriminant (t 1, t 2 ) = (x 1 x 2 ) 2 is not zero, where x i are roots of the polynomial in the algebraic closure F p (but NOT necessarily in F p ). GilYoung Cheong (University of Michigan) September 21, / 41

23 Ellenberg s argument Notice that we have (t 1, t 2 ) = (x 1 x 2 ) 2 = x 2 1 2x 1 x 2 + x 2 2 = (x 1 + x 2 ) 2 4x 1 x 2 = t 2 1 4t 2, so (t 1, t 2 ) is a polynomial expression in t 1, t 2, whose coefficients come from Z. GilYoung Cheong (University of Michigan) September 21, / 41

24 Ellenberg s argument Therefore, we have a bijection: monic, sq-free F p [x] deg=2 {(t 1, t 2 ) A 2 (F p ) : t1 2 4t 2} given by x 2 + t 1 x + t 2 (t 1, t 2 ). Thus, counting the elements of the left-hand side is same as counting that of the right-hand side. Why do we do this? Remark The right-hand side has some geometry! GilYoung Cheong (University of Michigan) September 21, / 41

25 Ellenberg s argument What do we mean by geometry? Replace the finite field F p with C: This means that we have monic, sq-free C[x] deg=2 {(t 1, t 2 ) C 2 : t1 2 4t 2}. where x 1, x 2 C are DISTINCT. x 2 + t 1 x + t 2 = (x x 1 )(x x 2 ) GilYoung Cheong (University of Michigan) September 21, / 41

26 Ellenberg s argument That is, we get a bijection {(t 1, t 2 ) C 2 : t 2 1 4t 2} {{x 1, x 2 } : x 1, x 2 C distinct} =: Conf 2 (C). The right-hand side is a well-studied space called the configuration space of 2 distinct points on the plane C = R 2! GilYoung Cheong (University of Michigan) September 21, / 41

27 Ellenberg s argument What Ellenberg noticed is that the Betti numbers gives us the count h 0 (Conf 2 (C)), h 1 (Conf 2 (C)), h 2 (Conf 2 (C)) #{(t 1, t 2 ) F p : t1 2 4t monic, sq-free 2} = #F p [x] deg=2. GilYoung Cheong (University of Michigan) September 21, / 41

28 Ellenberg s argument: What are the Betti numbers of a space? The Betti numbers h 0 (X ), h 1 (X ), h 2 (X ), of a space X store some essential topological information about the space. Here s an example. Given a surface X, we define the Euler characteristic of X by where v(x ) is the number of vertices; e(x ) is the number of edges; f (X ) is the number of faces. χ(x ) = v(x ) e(x ) + f (X ) Let s compute χ(x ) for the following examples. GilYoung Cheong (University of Michigan) September 21, / 41

29 Ellenberg s argument: Which information is cohomological information? (Excerpted from simomaths.files.wordpress.com) For any of the surface X above, we get χ(x ) = 2. Is this magic? GilYoung Cheong (University of Michigan) September 21, / 41

30 Ellenberg s argument We briefly sketch why this happens. For a surface X, denote h 0 (X ) the number of connected components (of X ); h 1 (X ) the number of 2-dimensional holes; h 2 (X ) the number of 3-dimensional holes. For a surface X, one can prove that χ(x ) = h 0 (X ) h 1 (X ) + h 2 (X ) in the first course of algebraic topology. In all previous examples, we had h 0 (X ) = h 1 (X ) = 1 and h 1 (X ) = 0, so χ(x ) = = 2, which is why we got the same answers. GilYoung Cheong (University of Michigan) September 21, / 41

31 Ellenberg s argument Back to our counting problem, in a framework of algebraic geometry, Ellenberg used a result of Minhyong Kim to establish the following formula monic, sq-free #F p [x] deg=2 = #{(t 1, t 2 ) A 2 (F p ) : t1 2 4t 4 } = i 0( 1) i p 2 i h i (Conf 2 (C)) = p 2 h 0 (Conf 2 (C)) ph 1 (Conf d (C)) + h 2 (Conf 2 (C)). GilYoung Cheong (University of Michigan) September 21, / 41

32 Ellenberg s argument Now, it remains to compute h i (Conf 2 (C)) for i 0. But already in 1970, Vladimir Arnol d proved the following. Theorem (Arnol d, 1970) if d 2, and if d = 0, 1. h i (Conf d (C)) = h i (Conf d (C)) = { 1 if i = 0, 1 0 if i 2 { 1 if i = 0 0 if i 1 GilYoung Cheong (University of Michigan) September 21, / 41

33 Ellenberg s argument Applying Arnol d s theorem for d = 2, we have monic, sq-free #F p [x] deg=2 = p 2 h 0 (Conf 2 (C)) ph 1 (Conf 2 (C)) + h 2 (Conf 2 (C)) = p 2 1 p = p 2 p, which is the same as what we got from Zieve s argument. GilYoung Cheong (University of Michigan) September 21, / 41

34 Exercise. Generalize Ellenberg s formula to the following: monic, sq-free #F p [x] deg=d = i 0( 1) i p d i h i (Conf d (C)) where p is a prime and d is any non-negative integer. Use Arnol d s result monic, sq-free to deduce that #F p [x] deg=d = p d p d 1 for d 2. Remark After you have done the exercises so far, you will realize that Zieve s monic, sq-free method to count #F p [x] deg=d and Ellenberg s formula together reproves Arnol d s result. GilYoung Cheong (University of Michigan) September 21, / 41

35 Thank you! (Part I) GilYoung Cheong (University of Michigan) September 21, / 41

36 Generalizations Around 2012, during an undergraduate research project, I realized that monic sq-free #{f F p [x] deg=d : f (0) 0} = (p 1)(p d ( 1) d )/(p + 1) = p d 2p d 1 + 2p d 2 + ( 1) d 1 2p + ( 1) d, for d 2. Does it remind you of Ellenberg s argument? Well, here is a topological result of Goryunov in GilYoung Cheong (University of Michigan) September 21, / 41

37 Generalizations Theorem (Goryunov, 1982) If d 2, we have 1 if i = 0, d h i (Conf d (C \ {0})) = 2 if 1 i d 1 0 otherwise. Thus, numerically speaking, what we have is monic sq-free #{f F p [x] deg=d : f (0) 0} = i 0 ( 1)i p d i h i (Conf d (C \ {0})) for any d 2. Wait, does this happen in even more generality? GilYoung Cheong (University of Michigan) September 21, / 41

38 Non-vanishing conditions = Puntures The answer is YES. Theorem (C-) Let p be any prime. Fix any distinct x 1,, x r F p, where 0 r p. We have monic sq-free {f F p [x] deg=d : f (x i ) 0 for 1 i r} = where d ( 1) i p d i h i (d, r), i=0 h i (d, r) = h i (Conf d (C \ {r points}) GilYoung Cheong (University of Michigan) September 21, / 41

39 Computation of the cohomological information Remark We note that the numbers {h i (d, r)} i,d,r 0 are already known by a work of Napolitano (2003) that h 0 (d, r) = 1 for any d, r 0; h i (d, r) = 0 whenever i > d; { 1 if d 2 h 1 (d, 0) = 0 if d = 0, 1 ; h i (d, r) = d j=0 hi j (d j, r 1), so the theorem has actual numerical meaning. This is in fact the key of the current proof of the theorem. GilYoung Cheong (University of Michigan) September 21, / 41

40 Further question Question Can we prove the theorem: monic sq-free #{f F p [x] deg=d : f (x i ) 0 for 1 i r} = without using Napolitano s numerical description? d ( 1) i p d i h i (d, r), i=0 Answering this is an ongoing project, and this is a direct generalization of Ellenberg s argument, which uses a more sophisticated technique in algebraic geometry, called étale cohomology. GilYoung Cheong (University of Michigan) September 21, / 41

41 Thank you! (Part II) GilYoung Cheong (University of Michigan) September 21, / 41

Polynomials. Chapter 4

Polynomials. Chapter 4 Chapter 4 Polynomials In this Chapter we shall see that everything we did with integers in the last Chapter we can also do with polynomials. Fix a field F (e.g. F = Q, R, C or Z/(p) for a prime p). Notation