EQUIAREAL TRIANGULATIONS AND P-ADIC VALUATION. Definition 1. For any field K, a function v : K ! R>0 2 Bon-Soon Lin (Updated Friday 31st July, 2020 at 10:02pm) is called a real valuation on K if it satisfies (1) Positive-definiteness: For every x 2 K, we have v(x) > 0 and v(x) = 0 if and only if x = 0 in K. Warm-up. (2) Multiplicative: For any x; y 2 K, we have v(xy) = v(x)v(y). (1) If a polygon P is triangulated into m many triangles, and that there are b (3) Subadditivity: For any x; y 2 K, we have v(x + y) 6 v(x) + v(y). many vertices on the boundary of P (including the ones made for the triangu- lation), and i many internal vertices. What is the relation between m, b, and i? We have the following immediate observations: 1 Exercise 2. For any real valuation v on field K: (2) Draw a square, can you triangulate it (with geometric triangles) such that (1) v(1) = 1. each triangle have the same area (the triangles need not be similar)? Such a tri- angulation is called an equiareal triangulation. Does there exist an equiareal (2) v(1=x) = 1=v(x) for all x 6= 0 in K. triangulation of a square with 6 triangles? How about with an even number of (3) v(−1) = 1. triangles? Here are some examples of real valuations: (3) Does there exist an equiareal triangulation of a square with 3 triangles? Or 5? (1) The usualJuly, 2020 absolute / norm function on Q, R, and C. (4) Does there exist an equiareal triangulation of a regular hexagon with 4 tri- (2) The trivial valuation on a field K, where v(x) = 1 whenever x 6= 0, and angles? Make some conjectures. v(0) = 0. st (5) Do all quadrilaterals have equiareal triangulations? That is, given a quadri- (3) The p-adic valuation on Q for a fixed prime p. Fix a prime p, define lateral Q, can you always triangulate it into some number of equal area triangles? vp : Q ! R>0 (6) Show it is impossible to have a triangulation of a square into 3 equal area by v (x) = 1 if x = pa n where gcd(n; p) = gcd(m; p) = 1for all x 6= 0, and triangles. p pa m vp(0) = 0. That is, factor out as many powers of the prime p from the numerator and the denominator of x, and then take its reciprocal. Here vp(x) measures 1. Real valuations and p-adic valuations. how many times you can divide p into x, and the more factors of p you can divide into x, the smaller vp(x) would get. This is consistent with vp(0) = 0, as Recall the absolute value function on the real numbers, you can divide however many copies of p into zero. |·| : ! ; R R>0 Exercise 3. What is v2(16)? v2(10001)? Make an observation here about the Prepared by Bon-Soon Lin on: 2-adic valuationFriday 31 v on . What about v (4) = 1=4 and v (16) = 1=16 and which satisfies the following three main properties: (1) For every x 2 R, we have 2 Q 2 2 v (20) = 1=4. jxj > 0 and jxj = 0 if and only if x = 0. (2) For any two x; y 2 R, we have 2 jxyj = jxj jyj. (3) For any two x; y 2 , we have jx + yj jxj + jyj. R 6 Exercise 4 (Divisibility criterion). Let m be an integer. If vp(m) = 1, what We make the following definition for any field K: can you say about m? If vp(m) < 1, what can you say about m? Can vp(m) > 1 for integer m? 1Hint: Recall Euler polyhedral formula: For any connected planar graph, we have V − E + F = 2 (including the outer face). 2sometimes people call it real norm or real absolute value instead 1 EQUIAREAL TRIANGULATIONS AND P-ADIC VALUATION. Bon-Soon Lin In fact, the p-adic valuations on Q satisfy a stronger subadditivity condition maxfv(x); 1g. Now consider v(x + 1)n and binomial expand, write an inequality i called non-Archimedean: For a field K and function v : K ! R>0 with v being a valuation and the given condition. And observe that v(x) 6 n (3’) Non-Archimedean: For all x; y 2 K, v(x + y) 6 max(v(x); v(y)). maxfv(x) ; 1g whenever 0 6 i 6 n.] Exercise 5. Prove indeed that the p-adic valuation v on is non-Archimedean.3 p Q 2. Extension of valuations. Exercise 6. Prove that if v : K ! is non-Archimedean then it is subaddi- R>0 We have seen the usual valuation functions on , and (namely the usual tive. But subadditivity does not imply non-Archimedean. Q R C absolute value function), and a non-Archimedean on Q, the p-adic valuation vp In fact, a non-Archimedean valuation satisfies a domination condition: for a fixed prime p. Sometimes we like to keep the features of vp on Q and extend it to all of R. And indeed we have the following theorem: Proposition 7 (Domination). If v : K ! R>0 is a non-Archimedean real valuation, then v(x + y) = max(v(x); v(y)) whenever v(x) 6= v(y). And in fact, Theorem 13 (Chevalley extension theorem). For any real valuation v : K ! and any field extension L=K, there exists a valuation v : L ! extending if for a; b1; b2; : : : ; bn 2 K we have v(a) > v(bj), then v(a + b1 + ··· + bn) = v(a). R>0 e R>0 v, namely vj = v, if either (1) L=K is algebraic, or (2) v is non-Archimedean. Exercise 8. Is the domination condition true for non-Archimedean valuations e K Furthermore, if v : K ! is a non-Archimedean valuation and t is a tran- if v(x) = v(y)? R>0 scendental element to the field K, then for any choice of positive number c > 0, Exercise 9. Show if v is a non-Archimedean valuation, then v(x + y) < v(y) there exists a valuation v : K(t) ! extending v such that v(t) = c. e R>0 e implies v(x) = v(y).4 We will not prove above theorem, but we will use it whenever needed. In par- What are some other non-Archimedean valuations on Q? As it turns out the ticular: p-adic ones are essentially the ones there are, up to some power factor: Exercise 14. Show there exists a real valuation v on R such that v(1=2) > 1. Exercise 10. Let v : Q ! R 0 be a non-trivial non-Archimedean valuation on July, 2020 > Exercise 15. Show for any non-Archimedean valuation v on such that v(2) < Q (so it does not just assign 1 to every nonzero q 2 Q). R 1, we have v(2k + 1) = 1, for every natural number k. Hint: What can you say (1) Show for each integer n 2 , we have v(n) 1. Z 6 about v(2k)? (2) Show there exists some prime p such that v(p) < 1. st (3) Show if q is a different prime from p, then v(q) = 1. 5 Exercise 16. For any prime p, and a 2 R any transcendental real number, the p-adic valuation vp on can be extended to vp on (a) such that vp(a) = 1. Is − logp k Q e Q e (4) Suppose v(p) = k, show v(x) = vp(x) for all x 2 Q. vep still non-Archimedean? Remark 11. Above is part of Ostrowski’s theorem: Two real valuations v ; v 1 2 Exercise 17. Consider the 2-adic valuation v on . Suppose we extend v to on field K are called equivalent if there exists some c > 0 such that v = vc. 2 Q 2 1 2 a valuation v on the field (π) such that v(π) = 4. What is v(π2)? What is For any non-trivial valuation v on , v is equivalent to the usual real absolute e Q e e Q v(π=4)? value or the usual p-adic valuation for some prime p. e Exercise 12 (Characterization of non-Archimedean valuations). Let v : K ! Bottom line: If v is a non-Archimedean valuation, we can always extend to a larger field in particular from Q to R. In addition we can, one at a time, assign R>0 be a real valuation on field K. Then v is non-Archimedean if and only if for all n 2 , v(n) 1.6 a transcendental element to any value that we want. N 6Prepared by Bon-Soon Lin on: Friday 31 [Hints: We need to show if v(n) 6 1 for all n 2 N, we have for all x; y 2 K, 3. Equiareal triangulations – the square. v(x + y) 6 maxfv(x); v(y)g. If y 6= 0, then it reduces to showing v(x + 1) 6 We now turn to our geometric question. Given a polygon P , can you triangulate 3Show the non-Archimedean property holds for integers first. P into m many triangles such that the triangles have the same area, for some 4Suppose v(x) 6= v(y), apply domination. 5Try Bezout’s theorem on GCD of two numbers, that there exists integers a; b such that m? If so, for which m? If indeed we can triangulate P into m many equal area ap + bq = 1.