Teaching Commutative Algebra and Algebraic Geometry. Geometry using Computer Algebra Systems

Transcription

1 Teaching Commutative Algebra and Algebraic Geometry using Computer Algebra Systems Department of Mathematics, Simon Fraser University British Columbia, CANADA Computer Algebra in Education ACA 2012, Sofia, Bulgaria June 25-28, 2012

2 Talk Outline MATH 441 Commutative Algebra and Algebraic Geometry Simon Fraser University, 2006, 2008, 2010, Who takes the course? Textbook and course content. Maple and Assessment. Three applications. Read material (paper). Reproduce computational results. Correct errors. Course project for graduate students.

3 Who takes the course? 4th undergraduate students and 1st year graduate students. major total mathematics computing math & cmpt other graduate total

4 Textbook Ch. Ch. Ch. Ch. Ch. Ch. Ch. Ch Geometry, Algebra and Algorithms Groebner Bases Elimination Theory The Algebra-Geometry Dictionary Quotient Rings Automatic Geometric Theorem Proving Invariant Theory of Finite Groups Projective Algebraic Geometry Appendix C. Computer Algebra Systems

5 Content 1 Varieties, Graphing varieties, Ideals. 2 Monomial orderings and the division algorithm. The Hilbert basis theorem. Gröbner bases and Buchberger s algorithm. 3 Solving equations (using Gröbner bases). Elimination theory and resultants. 4 Hilbert s Nullstellensatz. Radical ideals and radical membership. Zariski topology. Irreducible varieties, prime ideals, maximal ideals. Ideal decomposition. 5 Quotient rings, computing in quotient rings. 6 Applications (of Gröbner bases).

6 Maple and Assessment One intro Maple tutorial in lab. Detailed examples worksheet for self study. Five in class demos. Maple worksheet handouts.

7 Maple and Assessment One intro Maple tutorial in lab. Detailed examples worksheet for self study. Five in class demos. Maple worksheet handouts. MATH 441 MATH assignments 60% 60% project 10% 24 hour final 40% 30% Assignment questions, final exam, and project need Maple. Post take home final on web at 9am. Hand in following day before 10am in person.

8 Application 1: Circle Packing Problems Pack n = 6 circles in the unit square maximizing the radius r. Pack n = 6 points in the unit square maximizing their separating distance m. r = m 2(m + 1) D. Würtz, M. Monagan and R. Peikert. The History of Packing Circles in a Square. MapleTech, Birkhauser, 1994.

9 Application 1: Circle Packing Problems Given a packing, find m. Let P i = (x i, y i ) for 1 i 6. So P 1 = (0, 0), P 6 = (x 6, y 6 ), etc. Pythagoras: (x 6 x 1 ) 2 + (y 6 y 1 ) 2 = m 2. Symmetry: x 6 = 1/2, y 6 = (y 0 + y 5 )/2. Do not solve for m, x 1,..., x 6, y 1,..., y 6. Instead let

10 Application 1: Circle Packing Problems Given a packing, find m. Let P i = (x i, y i ) for 1 i 6. So P 1 = (0, 0), P 6 = (x 6, y 6 ), etc. Pythagoras: (x 6 x 1 ) 2 + (y 6 y 1 ) 2 = m 2. Symmetry: x 6 = 1/2, y 6 = (y 0 + y 5 )/2. Do not solve for m, x 1,..., x 6, y 1,..., y 6. Instead let I = x 6 1 2, x y 2 6 m 2,... Q[x 1,..., x 6, y 1,..., y 6, m] and compute a Gröbner basis G for I Q[m] = g. I get G = {(4m 2 5) (36m 2 13)}. Figure out that 4m 2 5 = 0 is a degenerate case.

11 Application 1: Circle Packing Problems Case n = 10 m = m = m = m = J. Schaer R. Milano G. Valette WMP What can go wrong?

12 Application 1: Circle Packing Problems Case n = 10 m = m = m = m = J. Schaer R. Milano G. Valette WMP What can go wrong? Input equations incorrectly.

13 Application 1: Circle Packing Problems Case n = 10 m = m = m = m = J. Schaer R. Milano G. Valette WMP What can go wrong? Input equations incorrectly. Errors in the figures.

14 Application 1: Circle Packing Problems Case n = 10 m = m = m = m = J. Schaer R. Milano G. Valette WMP What can go wrong? Input equations incorrectly. Errors in the figures. Setup may have degenerate solutions.

15 Application 1: Circle Packing Problems Case n = 10 m = m = m = m = J. Schaer R. Milano G. Valette WMP What can go wrong? Input equations incorrectly. Errors in the figures. Setup may have degenerate solutions. Too many quadratic equations = long time.

16 Application 2: Graph Coloring and Hilbert s Nullstellensatz. Which of these graphs can be colored with three colors? Figure: Wheel graphs W 3 and W

17 Application 2: Graph Coloring and Hilbert s Nullstellensatz. To color a graph on n vertices with k = 3 colors, set S := {x k 1 = 1, x k 2 = 1,..., x k n = 1}. For each edge (u, v) G set S := S { } x k u xv k = 0. x u x v

18 Application 2: Graph Coloring and Hilbert s Nullstellensatz. To color a graph on n vertices with k = 3 colors, set S := {x k 1 = 1, x k 2 = 1,..., x k n = 1}. For each edge (u, v) G set S := S Theorem G is k-colorable S has solutions over C. { } x k u xv k = 0. x u x v

19 Application 2: Graph Coloring and Hilbert s Nullstellensatz. To color a graph on n vertices with k = 3 colors, set S := {x k 1 = 1, x k 2 = 1,..., x k n = 1}. For each edge (u, v) G set S := S Theorem G is k-colorable S has solutions over C. { } x k u xv k = 0. x u x v Nice! So let I = x k 1 1,...,. Compute a reduced Gröbner basis B for I. Et voila! B = {1} G is not k-colorable.

20 Application 2: Graph Coloring and Hilbert s Nullstellensatz. But Theorem Graph k-colorability is NP-complete for k 3.

21 Application 2: Graph Coloring and Hilbert s Nullstellensatz. But Theorem Graph k-colorability is NP-complete for k 3. J.A. de Loera, J. Lee, P.N. Malkin and S. Margulies. Hilbert s Nullstellensatz and an algorithm for proving combinatorial infeasibility. In Proc. ISSAC 2008, ACM Press, , 2008.

22 Application 2: Graph Coloring and Hilbert s Nullstellensatz. Let V = V(x k 1 1,...,...) and I = x k 1 1,...,. Theorem G is NOT k colorable V = φ HNS 1 I.

23 Application 2: Graph Coloring and Hilbert s Nullstellensatz. Let V = V(x k 1 1,...,...) and I = x k 1 1,...,. Theorem G is NOT k colorable V = φ HNS 1 I. But if I = f 1, f 2,..., f m Q[x 1, x 2,..., x n ] then 1 I = 1 = h 1 f 1 + h 2 f h m f m for some h 1, h 2,..., h m in Q[x 1, x 2,..., x n ]. Idea 1: Try to find h i with degree d = 1, 2, 3,.... Idea 2: The larger d the harder the combinatorial problem. Idea 3: Replace Q with F 2. Get the students to experiment.

24 Application 3: Automatic Geometric Theorem Proving. Theorem C D N A B Let ABCD be a parallelogram and N = AC BD. Then N is the midpoint of AC and BD. Can we automate the proof?

25 Application 3: Automatic Geometric Theorem Proving. C = (u 2, u 3 ) D = (x 1, y 1 ) N = (x 2, y 2 ) A=(0,0) B = (u 1, 0) Step 2: Need 4 equations. Step 1: Fix co-ordinates. 3 parameters u 1, u 2, u 3. 4 unknowns x 1, y 1, x 2, y 2. Solutions are in R(u 1, u 2, u 3 ).

26 Application 3: Automatic Geometric Theorem Proving. C = (u 2, u 3 ) D = (x 1, y 1 ) N = (x 2, y 2 ) A=(0,0) B = (u 1, 0) Step 1: Fix co-ordinates. 3 parameters u 1, u 2, u 3. 4 unknowns x 1, y 1, x 2, y 2. Solutions are in R(u 1, u 2, u 3 ). Step 2: Need 4 equations. ABDC is a parallelogram = the slope of AC = BD = u 3 u 2 = y 1 x 1 u 1 = (x 1 u 1 )u 3 = u 2 y 1. Similarly the slope of AB = CD = y 1 = u 3.

27 Application 3: Automatic Geometric Theorem Proving. C = (u 2, u 3 ) D = (x 1, y 1 ) N = (x 2, y 2 ) A=(0,0) B = (u 1, 0) Let N be the intersection of AD and BC. Hence A, N, D are co-linear = [ ] x2 x det( 1 ) = 0 = x y 2 y 2 y 1 y 2 x 1 = 0. 1 Similarly B, N, C are co-linear = (u 1 u 2 )y 2 = u 3 (u 1 x 2 ).

28 Application 3: Automatic Geometric Theorem Proving. C = (u 2, u 3 ) D = (x 1, y 1 ) Equations N = (x 2, y 2 ) h 1 = (x 1 u 1 )u 3 u 2 y 1 h 2 = y 1 u 3 h 3 = x 2 y 1 y 2 x 1 h 4 = (u 1 u 2 )y 2 u 3 (u 1 x 2 ) A=(0,0) B = (u 1, 0) Step 2 (cont.): To prove N is the midpoint of AD and BC show N A 2 = D N 2 = x y 2 2 = (x 1 x 2 ) 2 + (y 1 y 2 ) 2. Similarly N B 2 = C N 2 = (x 2 u 1 ) 2 + y 2 2 = (u 2 x 2 ) 2 + u 2 3.

29 Application 3: Automatic Geometric Theorem Proving. h 1 = (x 1 u 1 )u 3 u 2 y 1 h 2 = y 1 u 3 h 3 = x 2 y 1 y 2 x 1 h 4 = (u 1 u 2 )y 2 u 3 (u 1 x 2 ) g 1 = x y 2 2 (x 1 x 2 ) 2 (y 1 y 2 ) 2 g 2 = (x 2 u 1 ) 2 + y 2 2 (u 2 x 2 ) 2 u 2 3 Step 3. Computation to prove theorem. Let I = h 1, h 2, h 3, h 4 R(u 1, u 2, u 3 )[x 1, y 1, x 2, y 2 ].

30 Application 3: Automatic Geometric Theorem Proving. h 1 = (x 1 u 1 )u 3 u 2 y 1 h 2 = y 1 u 3 h 3 = x 2 y 1 y 2 x 1 h 4 = (u 1 u 2 )y 2 u 3 (u 1 x 2 ) g 1 = x y 2 2 (x 1 x 2 ) 2 (y 1 y 2 ) 2 g 2 = (x 2 u 1 ) 2 + y 2 2 (u 2 x 2 ) 2 u 2 3 Step 3. Computation to prove theorem. Let I = h 1, h 2, h 3, h 4 R(u 1, u 2, u 3 )[x 1, y 1, x 2, y 2 ]. Then g 1 V(h 1, h 2, h 3, h 4 ) g 1 I 1 h 1, h 2, h 3, h 4, 1 g 1 z R(u 1, u 2, u 3 )[x 1, y 1, x 2, y 2, z]. Similarly verify g 2 I. What can go wrong?

31 Application 3: What can go wrong? Errors: any claim is true if I = h 1, h 2,... = 1. = check that the Gröbner basis for I is not {1}!!

32 Application 3: What can go wrong? Errors: any claim is true if I = h 1, h 2,... = 1. = check that the Gröbner basis for I is not {1}!! Show that working in R[u 1, u 1, u 3, x 1, y 1, x 2, y 2 ] leads to the degenerate cases u 1 = 0 where the theorem does not hold.

33 Application 3: What can go wrong? Errors: any claim is true if I = h 1, h 2,... = 1. = check that the Gröbner basis for I is not {1}!! Show that working in R[u 1, u 1, u 3, x 1, y 1, x 2, y 2 ] leads to the degenerate cases u 1 = 0 where the theorem does not hold. C = (u 2, u 3 ) D = (x 1, y 1 ) N = (x 2, y 2 ) A=(0,0) B = (u 1, 0) N is the midpoint of AD = N = (A+D) so 2 x 2 = x 1 2, y 2 = y 1!! 2

34 Graduate student project 1 Implement Buchberger s algorithm. 2 Study and implement the FGLM basis conversion. J.C. Faugere, P. Gianni, D. Lazard, T. Mora. Efficient computation of zero-dimensional Gröbner bases by change of ordering. J. Symb. Comp., 16, , Show that FGLM works using Trinks system. {45p + 35s 165b = 36, 35p + 40z + 25t 27s = 0, 15w +25ps +30z 18t 165b 2 = 0, 9w +15pt +20zs = 0, wp + 2zt = 11b 3, 99w 11sb + 3b 2 = 0}

35 Thank you for coming. On-line course materials