Finding Minimal Polynomials with a Norm Calculator


 Ezra Pope
 5 months ago
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1 Finding Minimal Polynomials with a Norm Calculator Winona State University October 18, 2008
2 Algebraic Review An algebraic number, ζ, is a root of an irreducible monic polynomial with rational coefficients.
3 Algebraic Review An algebraic number, ζ, is a root of an irreducible monic polynomial with rational coefficients. Examples ζ = ζ = i x = 0 x x
4 Algebraic Review An algebraic number, ζ, is a root of an irreducible monic polynomial with rational coefficients. Examples ζ = ζ = i x = 0 x x This polynomial, M ζ (x), is called the minimal polynomial and M ζ (x) = i (x σ i (ζ)) where σ i Gal(Q(ζ)/Q).
5 Algebraic Review An algebraic number, ζ, is a root of an irreducible monic polynomial with rational coefficients. Examples ζ = ζ = i x = 0 x x This polynomial, M ζ (x), is called the minimal polynomial and M ζ (x) = i (x σ i (ζ)) where σ i Gal(Q(ζ)/Q). The (absolute) norm of ζ, norm(ζ) = σ i (ζ), is the absolute value of the constant term of M ζ (x).
6 Problem Statement Given A collection of unknown algbraic numbers {ζ 1, ζ 2,... } whose minimal polynomials have known degrees {d 1, d 2,... }.
7 Problem Statement Given A collection of unknown algbraic numbers {ζ 1, ζ 2,... } whose minimal polynomials have known degrees {d 1, d 2,... }. An algorithm that can compute the norm of an unknown algebraic number (norm calculator).
8 Problem Statement Given A collection of unknown algbraic numbers {ζ 1, ζ 2,... } whose minimal polynomials have known degrees {d 1, d 2,... }. An algorithm that can compute the norm of an unknown algebraic number (norm calculator). Find The minimal polynomials for (at least some of) the ζ k s?
9 Shimura Curves The Shimura curve S 6 is a Riemannian surface of genus zero.
10 Shimura Curves The Shimura curve S 6 is a Riemannian surface of genus zero. An isomorphism J : S 6 P 1 exists.
11 Shimura Curves The Shimura curve S 6 is a Riemannian surface of genus zero. An isomorphism J : S 6 P 1 exists. It can be uniquely specified by choosing the three points that map to 0, 1, and.
12 Shimura Curves The Shimura curve S 6 is a Riemannian surface of genus zero. An isomorphism J : S 6 P 1 exists. It can be uniquely specified by choosing the three points that map to 0, 1, and. Due to the properties of Shimura curves, no formula exists for such a map.
13 CM Points on the Shimura Curve There is a collection of special points {s k } S 6 called complex multiplication (CM) points.
14 CM Points on the Shimura Curve There is a collection of special points {s k } S 6 called complex multiplication (CM) points. Three of these, s 3, s 4, and s 24, are really special, so specify J by (s 3, s 4, s 24 ) J (0, 1, ).
15 CM Points on the Shimura Curve There is a collection of special points {s k } S 6 called complex multiplication (CM) points. Three of these, s 3, s 4, and s 24, are really special, so specify J by (s 3, s 4, s 24 ) J (0, 1, ). Then J maps all CM points to algebraic numbers.
16 CM Points on the Shimura Curve There is a collection of special points {s k } S 6 called complex multiplication (CM) points. Three of these, s 3, s 4, and s 24, are really special, so specify J by (s 3, s 4, s 24 ) J (0, 1, ). Then J maps all CM points to algebraic numbers. There exists an algorithm that calculates norm(j(s k )) for s k a CM point. (Errthum, 2007)
17 CM Points on the Shimura Curve There is a collection of special points {s k } S 6 called complex multiplication (CM) points. Three of these, s 3, s 4, and s 24, are really special, so specify J by (s 3, s 4, s 24 ) J (0, 1, ). Then J maps all CM points to algebraic numbers. There exists an algorithm that calculates norm(j(s k )) for s k a CM point. (Errthum, 2007) Using genus theory of groups, we can calculate d k, the degree of M J(sk )(x).
18 CM Points on the Shimura Curve There is a collection of special points {s k } S 6 called complex multiplication (CM) points. Three of these, s 3, s 4, and s 24, are really special, so specify J by (s 3, s 4, s 24 ) J (0, 1, ). Then J maps all CM points to algebraic numbers. There exists an algorithm that calculates norm(j(s k )) for s k a CM point. (Errthum, 2007) Using genus theory of groups, we can calculate d k, the degree of M J(sk )(x). Back to the problem: Let {ζ k } = {J(s k )}.
19 The Picture
20 First Trick For a finite number of indices r, d r = 1, i.e. ζ r is a rational number.
21 First Trick For a finite number of indices r, d r = 1, i.e. ζ r is a rational number. norm(ζ r ) = ±ζ r.
22 First Trick For a finite number of indices r, d r = 1, i.e. ζ r is a rational number. norm(ζ r ) = ±ζ r. Use a new J 1 by taking (s 3, s 4, s 24 ) (1, 0, ) instead of (0, 1, ). J 1 (s) = 1 J(s).
23 First Trick Example For a finite number of indices r, d r = 1, i.e. ζ r is a rational number. norm(ζ r ) = ±ζ r. Use a new J 1 by taking (s 3, s 4, s 24 ) (1, 0, ) instead of (0, 1, ). J 1 (s) = 1 J(s). norm(j(s r )) = 4/5 and norm(1 J(s r )) = 9/5 ζ r = J(s r ) = 4/5
24 First Trick Example For a finite number of indices r, d r = 1, i.e. ζ r is a rational number. norm(ζ r ) = ±ζ r. Use a new J 1 by taking (s 3, s 4, s 24 ) (1, 0, ) instead of (0, 1, ). J 1 (s) = 1 J(s). norm(j(s r )) = 4/5 and norm(1 J(s r )) = 9/5 Can calculate all rational ζ r this way. ζ r = J(s r ) = 4/5
25 Second Trick Choose ζ k with d k small.
26 Second Trick Choose ζ k with d k small. For d k + 1 choices of r, specify J r (s) = ζ r J(s) by (s 3, s r, s 24 ) Jr (ζ r, 0, ).
27 Second Trick Choose ζ k with d k small. For d k + 1 choices of r, specify J r (s) = ζ r J(s) by (s 3, s r, s 24 ) Jr (ζ r, 0, ). norm(ζ r J(s k )) = σ i (ζ r J(s k )) i
28 Second Trick Choose ζ k with d k small. For d k + 1 choices of r, specify J r (s) = ζ r J(s) by (s 3, s r, s 24 ) Jr (ζ r, 0, ). norm(ζ r J(s k )) = σ i (ζ r J(s k )) i = σ i (ζ r ζ k ) i
29 Second Trick Choose ζ k with d k small. For d k + 1 choices of r, specify J r (s) = ζ r J(s) by (s 3, s r, s 24 ) Jr (ζ r, 0, ). norm(ζ r J(s k )) = σ i (ζ r J(s k )) i = σ i (ζ r ζ k ) i = (ζ r σ i (ζ k )) i
30 Second Trick Choose ζ k with d k small. For d k + 1 choices of r, specify J r (s) = ζ r J(s) by (s 3, s r, s 24 ) Jr (ζ r, 0, ). norm(ζ r J(s k )) = σ i (ζ r J(s k )) i = σ i (ζ r ζ k ) i = (ζ r σ i (ζ k )) i = M ζk (ζ r )
31 Brute Force So we know d k + 1 points on the curve y = M ζk (x).
32 Brute Force So we know d k + 1 points on the curve y = M ζk (x). If it wasn t for the absolute value, we could use a standard polynomial fit and be done.
33 Brute Force So we know d k + 1 points on the curve y = M ζk (x). If it wasn t for the absolute value, we could use a standard polynomial fit and be done. Go through the 2 d k combinations of minus signs on the values until you find a monic polynomial. (There s only ever one. Proof?)
34 Brute Force So we know d k + 1 points on the curve y = M ζk (x). If it wasn t for the absolute value, we could use a standard polynomial fit and be done. Go through the 2 d k combinations of minus signs on the values until you find a monic polynomial. (There s only ever one. Proof?) If there are R rational ζ r, then we can use this method to find the minimal polynomial of any ζ k with d k R 2.
35 Example norm(ζ) = and d = 3.
36 Example norm(ζ) = and d = 3. We use the four CM points that map to 0, 1, 4 5, and 2 3 to find the data points:
37 Example norm(ζ) = and d = 3. We use the four CM points that map to 0, 1, 4 5, and 2 3 to find the data points: (0, norm(ζ)) = ( 0, 10 ) 17 (1, norm (1 ζ)) = ( 1, 25 ) ( , norm ( 4 5 ζ)) = ( 4 ( 2 3, norm ( 2 3 ζ)) = ( 2 3, 104 ) 459 5, )
38 Example Possible minimal polynomials: norm(ζ) = and d = 3. We use the four CM points that map to 0, 1, 4 5, and 2 3 to find the data points: (0, norm(ζ)) = ( 0, 10 ) 17 (1, norm (1 ζ)) = ( 1, 25 ) ( , norm ( 4 5 ζ)) = ( 4 ( 2 3, norm ( 2 3 ζ)) = ( 2 3, 104 ) 459 5, ) y = y = y = y = y = y = y = y = x x x x x x x x x x 1 2 x2 + x x x x x x x x x x x x x3
39 Example Possible minimal polynomials: norm(ζ) = and d = 3. We use the four CM points that map to 0, 1, 4 5, and 2 3 to find the data points: (0, norm(ζ)) = ( 0, 10 ) 17 (1, norm (1 ζ)) = ( 1, 25 ) ( , norm ( 4 5 ζ)) = ( 4 ( 2 3, norm ( 2 3 ζ)) = ( 2 3, 104 ) 459 5, ) y = y = y = y = y = y = y = y = x x x x x x x x x x 1 2 x2 + x x x x x x x x x x x x x3 So the minimal polynomial of ζ is M ζ (x) = x 1 2 x 2 + x 3.
40 Thanks Questions?