The document discusses simulating the construction of finite fields using Maple software. It presents Maple code that constructs the finite field F64 as the quotient ring F2[x]/(f(x)) where f(x) is the primitive polynomial x6 + x + 1. It also constructs the subfield F8 of F64. The code is intended to help students visualize finite fields before learning the theoretical foundations of the topic.

1. SIMULATING THE CONSTRUCTIONS OF
FINITE FIELDS USING MAPLETS
LOEKY HARYANTO
Mathematics Department, Hasanuddin University,
email: L.Haryanto@unhas.ac.id
HaryantoL@outlook.com
GSM#s: +6281342127598
Related presentations (will be uploaded soon):
Factorization of xN  1 over Fp

2. 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.

3. Theoretical Review
Given a prime p and a polynomial f(x) Fp[x] of degree m. Let q =
pm. We need f(x) to be primitive; i.e. it has a primitive root a that
generates the following multiplicative group of order N = pm  1
* ={1, a, a2, …, aN1}.
Fq
If a is primitive, then using the element 0  f (a), the (additive)
factor group Fp[x]/(f(x)) and with the obvious multiplicative
operator, we can construct a field by identifying the isomorphism
Fp[x]/(f(x))  Fq = Fq
*  {0} = {0, 1, a, a2, …, aN1}.
Main reference:
Chapter 3 of W. C. Huffman, V. Pless, Fundamentals of Error-
Correcting Codes, Cambridge Univ. Press, 2003

4. How Maplet determines if Fp[x]/(f(x))  Fq or Fp[x]/(g(x))  Fq?
Compute the order of the quotient rings! (Should be equal to pm)
Is q1 = |Fp[x]/(f(x))| = pm? Is q2 = |Fp[x]/(g(x))| = pm?

5. Here F2[x]/(f(x)) ≇ F32 and F2[x]/(g(x))  F16

6. Wait, CONFUSING NOTATIONS FOR NEW LEARNERS:
Different notations for the same mathematical object:
1. Fp or GF(p) or Zp are three different notations for the same (prime)
field; where p is prime and
Fp = {0, 1, …, p  1}.
2. Fq or GF(q) are two different notations for the same field; the field
Fq = {0, 1, a, a2 …, aq2} = Fp[x]/(f(x))
where f is primitive and of degree m, q = pm. For every k  m, the ak
can be presented as a polynomial of degree < m in the indeterminate a.
When N = pm  1, we have aq2 = aN1.
3. More confusing for a new learner is the identification between the
field Fq and its associate linear space:
Fq = Fp  Fp  …  Fp
where the right hand side consists of m factors.

7. 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 = pm, N = q – 1 = pm  1.
The intended constructed finite field of characteristic p is
Fq (or GF(q) = GF(pm))
The cyclic group is
<a> = {1, a, a2, …, aN1} = Fq
* = Fq
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

16. Notice that
a16 = a1.

17. Notice that
a18 = a3.

18. Notice that
a20 = a5.

19. Notice that
a22 = a7.

20. Notice that
a24 = a9.

21. Notice that
a26 = a11.

22. Notice that
a28 = a13.

23. Notice that
a30 = a15.

24. A little bit of finite field’s theory:
THE SUBFIELD Fs OF THE FIELD Fq
where q = pm and s = pr.
Here, Fq is the quotient ring F2[x]/<f(x)> where f(x) = x6 + x + 1.
THEOREM (Huffman, Pless, Th. 3.5.3 (modified)):
When q = pm and s = pr
(i) Fq has subfield Fs if and only if r | m;
(ii) if r | m, then there is only one field of order s, which is Fs, of the field Fq
The Maplets make use p = 2, q = 64 and s = 8 (equivalently, m = 6 and r = 3)
The constructed finite field of order 26 (including its elements) is
F64 (or GF(64))
The constructed subfield of order 23 (including its elements) is
F8 (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

25. a0 = 1,
b = a9,
b0 = 1
or
a0 = 1,
b = a4+a3
b0 = 1
* = <a>  < a9> = F8
F64
*

26. a1 = a,
b = a9,
b1 = a9
or
a1 = a,
b = a4 + a3,
b1 = a4 + a3,
* = <a>  < a9> = F8
F64
*

27. * = <a>  < a9> = F8
F64
*
a2 = a2,
b = a9,
b2 = a18
or
a2 = a2,
b = a4 + a3,
b2 = a3 +a2
+ a1 + 1

28. a3 = a3,
b = a9,
b3 = a27
or
a3 = a3,
b = a4+a3
b3 = a3+ a2
+ a
* = <a>  < a9> = F8
F64
*

29. a4 = a4,
b = a9,
b4 = a36
or
a4 = a4,
b = a4+a3
b4 = a4+ a2
+ a
* = <a>  < a9> = F8
F64
*

30. a5 = a5,
b = a9,
b5 = a45
or
a5 = a5,
b = a4+a3
b5 = a4+ a3
+ 1
* = <a>  < a9> = F8
F64
*

31. a6 = a6,
b = a9,
b6 = a54
or
a6 = a + 1
b = a4 + a3
b6 = a4 + a2
+ a + 1
* = <a>  < a9> = F8
F64
*

32. a7 = a7,
b = a9,
b7 = a63
or
a7 = a2 + a
b = a4 + a3
b7 = 1
* = <a>  < a9> = F8
F64
*

33. a8 = a8,
b = a9,
b8 = a72
or
a8 = a3 + a2
b = a4 + a3
b8 = a4 + a3
* = <a>  < a9> = F8
F64
*

34. a9 = a9,
b = a9,
b9 = a81
or
a9 = a4 + a3
b = a4 + a3
b9 = a3 + a2
+ a + 1
* = <a>  < a9> = F8
F64
*

35. a10 = a10,
b = a9,
b10 = a90
or
a10 = a5 + a4
b = a4 + a3
b10 = a3 + a2
+ a
* = <a>  < a9> = F8
F64
*

36. a11 = a11,
b = a9,
b11 = a99
or
a11 = a5 + a
+ 1
b = a4 + a3
b11 = a4 +
a2 + a
* = <a>  < a9> = F8
F64
*

37. a12 = a12,
b = a9,
b12 = a108
or
a12 = a2 + 1
b = a4 + a3
b12 = a4 +
a3 + 1
* = <a>  < a9> = F8
F64
*

38. a13 = a13,
b = a9,
b13 = a117
or
a13 = a3 + a
b = a4 + a3
b13 = a4 +
a2 + a + 1
* = <a>  < a9> = F8
F64
*

39. a14 = a14,
b = a9,
b14 = a126
or
a14 = a4 + a2
b = a4 + a3
b14 = 1
* = <a>  < a9> = F8
F64
*

40. a61 = a61,
b = a9,
b61 = a549
or
a61 = a5 + a4
+ 1
b = a4 + a3
b61 = a4 + a3
+ 1
* = <a>  < a9> = F8
F64
*

41. a62 = a62,
b = a9,
b62 = a558
or
a62 = a5 + 1
b = a4 + a3
b61 = a4 + a2
+ a + 1
* = <a>  < a9> = F8
F64
*

42. a63 = a63,
b = a9,
b63 = a567
or
a63 = 1
b = a4 + a3
b63 = 1
* = <a>  < a9> = F8
F64
*

43. Conclusion