SIMULATING THE CONSTRUCTIONS OF FINITE FIELDS USING MAPLETS L OEKY H ARYANTO Mathematics Department, Hasanuddin University, GSM#s: Related presentations (will be uploaded soon): Factorization of x N  1 over F p

A MOTIVATION FOR EVERY ABSTRACT ALGEBRA INSTRUCTOR: USE THIS PRESENTATION AS A NEW STRATEGY FOR STUDENT-CENTERED LEARNING (SCL) METHOD. The Maplet copies here were created to make students firstly being familiar with (not necessarily mastering the theory of) finite fields before the students being introduced with the theoretical parts of the subject; e.g. before they were given some formal theories which were written in the next page! By the way, since mathematics is a language which is full of written symbols, without visual and ‘seemingly’ interactive presentations, most of students tend to sleep in abstract algebra classes. Nevertheless, IMO most strategies proposed for the SCL method by experts in education are not appropriate for math classes, or even worse than the common usual (old) teaching method.

Theoretical Review Given a prime p and a polynomial f(x)  F p [x] of degree m. Let q = p m. We need f(x) to be primitive; i.e. it has a primitive root a that generates the following multiplicative group of order N = p m  1 F q * ={1, a, a 2, …, a N  1 }. If a is primitive, then using the element 0  f (a), the (additive) factor group F p [x]/(f(x)) and with the obvious multiplicative operator, we can construct a field by identifying the isomorphism F p [x]/(f(x))  F q = F q *  {0} = {0, 1, a, a 2, …, a N  1 }. Main reference: Chapter 3 of W. C. Huffman, V. Pless, Fundamentals of Error- Correcting Codes, Cambridge Univ. Press, 2003

How Maplet determines if F p [x]/(f(x))  F q or F p [x]/(g(x))  F q ? Compute the order of the quotient rings! (Should be equal to p m ) Is q 1 = |F p [x]/(f(x))| = p m ? Is q 2 = |F p [x]/(g(x))| = p m ?

Here F 2 [x]/(f(x)) ≇ F 32 and F 2 [x]/(g(x))  F 16

Wait, CONFUSING NOTATIONS FOR NEW LEARNERS: Different notations for the same mathematical object: 1. F p or GF(p) or Z p are three different notations for the same (prime) field; where p is prime and F p = {0, 1, …, p  1}. 2. F q or GF(q) are two different notations for the same field; the field F q = {0, 1, a, a 2 …, a q  2 } = F p [x]/(f(x)) where f is primitive and of degree m, q = p m. For every k  m, the a k can be presented as a polynomial of degree < m in the indeterminate a. When N = p m  1, we have a q  2 = a N  More confusing for a new learner is the identification between the field F q and its associate linear space: F q = F p  F p  …  F p where the right hand side consists of m factors.

A little bit of group theory: A CYCLIC GROUP GENERATED BY A ZERO OF A PRIMITIVE POLYNOMIAL f(x) OF DEGREE m. The zero of f(x) is a, i.e. f(a) = 0. Symbols: q = p m, N = q – 1 = p m  1. The intended constructed finite field of characteristic p is F q (or GF(q) = GF(p m )) The cyclic group is = {1, a, a 2, …, a N  1 } = F q * = F q DO NOT TRY TO MEMORIZE ALL THESE SYMBOLS RIGHT NOW. YOU WILL REMEMBER MOST OF THEM ONCE YOUR INSTRUCTOR KEEPS RUNNING AND EXPLAINING THE MATERIAL IN THIS PRESENTATION

Notice that a 16 = a 1.

Notice that a 18 = a 3.

Notice that a 20 = a 5.

Notice that a 22 = a 7.

Notice that a 24 = a 9.

Notice that a 26 = a 11.

Notice that a 28 = a 13.

Notice that a 30 = a 15.

A little bit of finite field’s theory: THE SUBFIELD F s OF THE FIELD F q where q = p m and s = p r. Here, F q is the quotient ring F 2 [x]/ where f(x) = x 6 + x + 1. THEOREM (Huffman, Pless, Th (modified)): When q = p m and s = p r (i) F q has subfield F s if and only if r | m; (ii) if r | m, then there is only one field of order s, which is F s, of the field F q The Maplets make use p = 2, q = 64 and s = 8 (equivalently, m = 6 and r = 3) The constructed finite field of order 2 6 (including its elements) is F 64 (or GF(64)) The constructed subfield of order 2 3 (including its elements) is F 8 (or GF(8)) DO NOT TRY TO MEMORIZE THESE THEORIES RIGHT NOW. YOU WILL REMEMBER MOST OF THEM ONCE YOUR INSTRUCTOR KEEPS RUNNING AND EXPLAINING THE MATERIAL IN THIS PRESENTATION

a 0 = 1, b = a 9, b 0 = 1 or a 0 = 1, b = a 4 +a 3 b 0 = 1 F 64 * =  = F 8 *

a 1 = a, b = a 9, b 1 = a 9 or a 1 = a, b = a 4 + a 3, b 1 = a 4 + a 3, F 64 * =  = F 8 *

a 2 = a 2, b = a 9, b 2 = a 18 or a 2 = a 2, b = a 4 + a 3, b 2 = a 3 +a 2 + a 1 + 1

a 3 = a 3, b = a 9, b 3 = a 27 or a 3 = a 3, b = a 4 +a 3 b 3 = a 3 + a 2 + a F 64 * =  = F 8 *

a 4 = a 4, b = a 9, b 4 = a 36 or a 4 = a 4, b = a 4 +a 3 b 4 = a 4 + a 2 + a F 64 * =  = F 8 *

a 5 = a 5, b = a 9, b 5 = a 45 or a 5 = a 5, b = a 4 +a 3 b 5 = a 4 + a F 64 * =  = F 8 *

a 6 = a 6, b = a 9, b 6 = a 54 or a 6 = a + 1 b = a 4 + a 3 b 6 = a 4 + a 2 + a + 1 F 64 * =  = F 8 *

a 7 = a 7, b = a 9, b 7 = a 63 or a 7 = a 2 + a b = a 4 + a 3 b 7 = 1 F 64 * =  = F 8 *

a 8 = a 8, b = a 9, b 8 = a 72 or a 8 = a 3 + a 2 b = a 4 + a 3 b 8 = a 4 + a 3 F 64 * =  = F 8 *

a 9 = a 9, b = a 9, b 9 = a 81 or a 9 = a 4 + a 3 b = a 4 + a 3 b 9 = a 3 + a 2 + a + 1 F 64 * =  = F 8 *

a 10 = a 10, b = a 9, b 10 = a 90 or a 10 = a 5 + a 4 b = a 4 + a 3 b 10 = a 3 + a 2 + a F 64 * =  = F 8 *

a 11 = a 11, b = a 9, b 11 = a 99 or a 11 = a 5 + a + 1 b = a 4 + a 3 b 11 = a 4 + a 2 + a F 64 * =  = F 8 *

a 12 = a 12, b = a 9, b 12 = a 108 or a 12 = a b = a 4 + a 3 b 12 = a 4 + a F 64 * =  = F 8 *

a 13 = a 13, b = a 9, b 13 = a 117 or a 13 = a 3 + a b = a 4 + a 3 b 13 = a 4 + a 2 + a + 1 F 64 * =  = F 8 *

a 14 = a 14, b = a 9, b 14 = a 126 or a 14 = a 4 + a 2 b = a 4 + a 3 b 14 = 1 F 64 * =  = F 8 *

a 61 = a 61, b = a 9, b 61 = a 549 or a 61 = a 5 + a b = a 4 + a 3 b 61 = a 4 + a F 64 * =  = F 8 *

a 62 = a 62, b = a 9, b 62 = a 558 or a 62 = a b = a 4 + a 3 b 61 = a 4 + a 2 + a + 1 F 64 * =  = F 8 *

a 63 = a 63, b = a 9, b 63 = a 567 or a 63 = 1 b = a 4 + a 3 b 63 = 1 F 64 * =  = F 8 *

Conclusion