Finite mathematics
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This document provides an overview of the topics that will be covered in a finite mathematics course, including residue arithmetic, elements of finite groups/rings/fields, number theory concepts like the Euclidean algorithm and Chinese Remainder Theorem, and basics of finite vector spaces and fields. The style of the course will be leisurely and discursive, focusing on mathematical thinking and discovery. While the mathematics is classical, it will be new to students. The goal is to emphasize elegance and aesthetics over utility alone.
Finite mathematics
- 1. FINITE MATHEMATICS Igor Rivin, St Andrews, Fall 2015
- 2. WHAT IS THIS ABOUT? • We will be learning about some basic basic techniques, which include: • Residue arithmetic • Elements of finite groups • Elements of finite rings and fields. One of the goals is proving Wedderburn’s theorem, which states that every finite division ring is a field. • Elements of number theory (Extended Euclidean algorithm, Chinese Remainder Theorem, etc) • Elements of cyclotomic polynomials. • The very basics of finite vector spaces, as well as affine and projective spaces.
- 3. WHAT IS THIS ABOUT, CONTINUED • The style of the course is leisurely and discourcive – we will take interesting diversions where we find them,The main point of the course is learning how (some) mathematicians think, and how we discover mathematics.While much of the mathematics we are covering is quite classical, all (or most) of it is new to us. • Our main criterion is not some putative utility, but aesthetics and elegance. One of the wonderful things about mathematics, is that beautiful things wind up being more useful! • What this means is that we often don’t know where we are going, until we get there, and these notes will be trailing behind the actual course a lot of the time.
- 4. RESIDUES • Consider an integer n. The set 𝑛𝑛ℤ is the set of multiples of n,so 𝑛𝑛ℤ = {𝑛𝑛, 2𝑛𝑛, 3𝑛𝑛, … }. We define an equivalence relation on the integers ℤ, by saying that 𝑎𝑎 ≡ 𝑏𝑏 (𝑚𝑚𝑚𝑚𝑚𝑚 𝑛𝑛), whenever 𝑎𝑎 − 𝑏𝑏 ∈ 𝑛𝑛𝑛.We denote the set of equivalence classes of this relation by ℤ/𝑛𝑛𝑛. • Given two equivalence classes, we can add and multiply them (by taking integer representatives, adding or multiplying those, and taking the equivalence class of the sum or product, respectively). • It is not hard to see that the classes of 0 and 1 are the additive and multiplicative identities, respectively, in ℤ/𝑛𝑛𝑛.
- 5. RESIDUES • We thus see that ℤ/𝑛𝑛𝑛 is a commutative ring with 1. (look up the definition!) • Recall that such a ring is called an integral domain if no non-zero element a is a divisor of zero. In other words, given a, there is a b, such that a b = 0, if and only if a=0. • Observation: ℤ/𝑛𝑛𝑛 is not an integral domain unless n is prime. Why? If n is not prime, then there are 1<k, l<n, such that n = k l. Clearly the residue classes of k and l are 0- divisors. • Similarly, we see that ℤ/𝑝𝑝ℤ, for p prime is an integral domain (if k is not a multiple of p, and l is not a multiple of p, we can’t have k l a multiple of p by the fundamental theorem of arithmetic.
- 6. FUNDAMENTAL THEOREM OF ARITHMETIC • Any positive integer n can be written as 𝑛𝑛 = 𝑝𝑝1 𝑎𝑎 𝑝𝑝2 𝑏𝑏 …𝑝𝑝𝑘𝑘 𝑐𝑐 , where the 𝑝𝑝𝑖𝑖 are prime, and this representation is unique up to the order of the factors. • Proof: existence is easy by induction: either n is prime, in which case there is nothing to do, or n = k l, in which case represent k and l by induction.The harder part is uniqueness, which we leave as a challenging problem (we will need the method we develop in the next little while.
- 7. RESIDUE RINGS MODULO A PRIME • For a prime p, we can show that ℤ/𝑝𝑝𝑝 is not just an integral domain, but a field. This means that for any nonzero a, there exists a b,such that 𝑎𝑎 𝑏𝑏 ≡ 1 𝑚𝑚𝑚𝑚𝑚𝑚 𝑝𝑝 . • PROOF 1: Consider the map 𝑀𝑀 𝑎𝑎 : ℤ/𝑝𝑝𝑝 ⟶ ℤ/𝑝𝑝𝑝 , given by M(a)(b) = a b. Since we know that ℤ/𝑝𝑝𝑝 is an integral domain, this map is 1-1, but a 1-1 map of finite sets of the same cardinality is a bijection, so there is a b, such that M(a)(b) = 1 = ab. • The above proof has the virtue of extreme simplicity, but the downside of being non- constructive (we know that the inverse exists, but don’t really know how to find one). • PROOF 2: This uses the fundamental properties of the Euclidean Algorithm (see the sequel) to show the following fundamental result:
- 8. CHARACTERIZATION OF THE GREATEST COMMON DIVISOR • Suppose m, n are integers.Then, the greatest common divisor of m, n is the smallest positive value of a linear combination with integer coefficients a m + b n. • The proof uses the Euclidean Algorithm, which is the following (essentially optimal) • We want to find the greatest common divisor of n and m. • Divide with remainder to write n = q m + r. • Note that the greatest common divisor of n and m is the same as the greatest common divisor of m and r, but note also that r < min(n,m). • Which means that the algorithm terminates.
- 9. EUCLIDEAN ALGORITHM • What is more, the remainder r is an integer linear combination of m and n. The next remainder will be a linear combination of m and r, and so of m and n. So will the last remainder (which is the gcd).That shows that the value of the smallest linear positive linear combination of m and n is no bigger than the gcd of m and n. But obviously, the gcd has to divide any linear combination, so we are done…
- 10. APPLICATIONS • ℤ/𝑝𝑝𝑝 is a field. Indeed, any n not divisible by a prime p is relatively prime to it, so there exist a, b such that a n + b p = 1. Taking equivalence classes modulo p of both sides, get a𝑛𝑛 ≡ 1 (𝑚𝑚𝑚𝑚𝑚𝑚 𝑝𝑝) • Chinese remainder theorem: given moduli 𝑛𝑛1, 𝑛𝑛2,… 𝑛𝑛𝑘𝑘, pairwise relatively prime, and remainders 𝑟𝑟1, 𝑟𝑟2,… 𝑟𝑟𝑘𝑘, there exists an integer x, such that x ≡ 𝑟𝑟𝑖𝑖 (mod 𝑛𝑛𝑖𝑖), for every i. • Proof of CRT by induction on k: first, do it for k=2. Look for x in the form • 𝑥𝑥 = 𝑎𝑎 𝑛𝑛1+b 𝑛𝑛2.We see that 𝑎𝑎𝑛𝑛1 ≡ 𝑟𝑟2 (𝑚𝑚𝑚𝑚𝑚𝑚 𝑛𝑛2), and 𝑏𝑏𝑛𝑛2 ≡ 𝑟𝑟1 (𝑚𝑚𝑚𝑚𝑚𝑚 𝑛𝑛1), so we see that 𝑎𝑎 ≡ 𝑟𝑟2 𝑛𝑛1 −1 (𝑚𝑚𝑚𝑚𝑚𝑚 𝑛𝑛2), and 𝑏𝑏 ≡ 𝑟𝑟1 𝑛𝑛2 −1 (𝑚𝑚𝑚𝑚𝑚𝑚 𝑛𝑛1).
- 11. FIELDS
- 12. CHARACTERISTIC OF A FIELD • The characteristic of a field F is the smallest integer c such that c x = 0, for any element x of F. If there is no such positive integer, the characteristic is said to be equal to 0. • Observation: the characteristic, if not equal to 0, has to be prime. If not, c = a b, so 0 = 𝑐𝑐 × 1 = 𝑎𝑎 × 1 × 𝑏𝑏 × 1 , which contradicts the existence of 0-divisors in a field. • Another observation: any field F of characteristic p contains a copy of 𝔽𝔽𝑝𝑝=ℤ/𝑝𝑝𝑝. Indeed, the elements 0,1, …, p-1 are such a copy.This copy is called the prime field of F.
- 13. VECTOR SPACES • A vector space V over a field F is an abelian group (which we will denote additively) with an action by F, satisfying: • 0x = 0, for all x in V. • 1x = x, for all x in V. • 𝑐𝑐 𝑑𝑑 𝑥𝑥 = 𝑐𝑐𝑐𝑐 𝑥𝑥, for all x in V, and c,d in F. • 𝑐𝑐 + 𝑑𝑑 𝑥𝑥 = 𝑐𝑐 𝑥𝑥 + 𝑑𝑑 𝑥𝑥, for all x in V, c, d in F. • 𝑐𝑐 𝑥𝑥 + 𝑦𝑦 = 𝑐𝑐 𝑥𝑥 + 𝑐𝑐 𝑦𝑦, for all c in F, and x, y in F.
- 14. VECTOR SPACES • A set of elements S= 𝑣𝑣1, 𝑣𝑣2 … , 𝑣𝑣𝑘𝑘 has a span, which is the set of all linear combinations of elements in S. This is denoted by 𝑆𝑆 . A set S is called spanning, if 𝑆𝑆 =V. • A set of elements S is called independent, if 𝑓𝑓1 𝑣𝑣1 + 𝑓𝑓2 𝑣𝑣2 + ⋯ 𝑓𝑓𝑘𝑘 𝑣𝑣𝑘𝑘 = 0 impliest that all of the 𝑓𝑓𝑖𝑖 are zero. • A set of elements S is called a basis, if it is both spanning and independent. • Every finite vector space V has a basis (keep adding elements) B, and the cardinality 𝑉𝑉 = 𝐹𝐹 𝐵𝐵
- 15. VECTOR SPACES AND FINITE FIELDS • Note that every finite field is a vector space over its prime field, so we see that • The cardinality of every finite field is 𝒑𝒑𝒌𝒌 , for some prime p. • Interesting fact: up to isomorphism, there is exactly one field of given cardinality.