應用代數學簡介

應用代數學簡介 2011. Sep. 李安莉於輔大數學系

近世代數學 • 近世代數是相對於古典代數來說的! • 古典代數基本上就是方程論,以探討方程式的解為中心的. • 隨方程論的進展, 複數系的發現使得方程式求解的問題跨進了一大步! 而一元方程式的根式解之研究引導出群的概念;漸漸的, 方程式論不再是代數學的全部,它漸漸地轉向對代數結構的本身研究!

方程式論 大學: Galois 理論群論 • 高中: • 代數基本定理 • 根與係數的關係 • Lagrange 內插公式 • Newton 近似根求法 • Strum 法則 (實係數多項式之根的分布情形) • … Galois Q1: 給一元方程式,有無解存在? Q2: 有無根式解存在? 在多數場合下,代數理論指研究代數運算本身之性質,而不管施行對象之具體屬性…這種新觀點是20世紀才產生的, 德國女數學家Noether在這方面有相當大的貢獻.

Algebra代數一詞來自阿拉伯文 al – jabr 通常的解釋是類似隱含在翻譯上面。這個詞的 al – jabr 大概意味著什麼“復辟”或 “完成”，似乎是指移位減去條件對方的方程 ;

關於這學年的代數學 • 上課時間: 每週二第3, 4堂 • 上課地點: MA301 • 評量方式: 期中考 30% 期末考 30% 平時考 40% 教學網站: http://www.math.fju.edu.tw/blog/index.php?PID=44

Part 1. Integers and Equivalence Relations • Chapter 0. Preliminaries: 1. Proofs 2. Sets 3. Mappings 4. Equivalences ◎等同學拿到課本後,請自行復習之!

4. Equivalence Relations • Def.An equivalence relation on a set S is a set R of ordered pairs of elements of S such that 1. (a,a) lies in R for all a in S (reflexive) 2. (a,b) lies in R implies (b,a) lies in R ( symmetric) 3. (a,b) lies in R and (b,c) lies in R implies (a,c) lies in R (transitive) Equivalence Classes Partition: The equivalence classes of an equivalence relation on a set S constitute a partition of S( a collection of nonempty disjoint subsets of S whose union is S) For every partition P of S, there is an equivalence relation on S whose equivalence classes are the elements of P.

Chapter 1. Integers and permutations • 1.1. Induction (略) • 1.2. Divisors and Prime Factorization • 1.3. Integers Modulo n • 1.4. Permutations • 1.5.Application to Cryptography

Sec.1.1. Induction • Well-Ordering Axiom: Every nonempty set of nonnegative integers has a smallest member. • The Principle of Induction: If m is any integer. Let P(m), P(m+1),…be statements such that (1). P(m) is true, (2). P(k)  P(k+1) for all k  m Then P(n) is true for all n  m. • Strong Induction: If m is any integer. Let P(m), P(m+1),…be statements such that (1). P(m) is true, (2).If k  m and P(m),P(m+1),…,P(k) are true  P(k+1) is also true Then P(n) is true for all n  m. Well-Ordering  Induction  Strong Induction Exercises!

1. Properties of Integers • Well Ordering Principle: Every nonempty set of integers contains a smallest member. • Division Algorithm: Let a and b be integers with b>0. Then there exist unique integers q and r with the property that a=b·q+r, where 0≤r<b. • GCD: • LCM: • Fundamental Theorem of Arithmetic: Every integer n >1 is a prime or a product of primes. This product is uniquely determined up to the orders.

2. Modular Arithmetic • Notation: a=q·b+r iff b| a-r iff a≡r mod b. • Some Basic Properties: Let n >0. 1. a ≡ a mod n 2. If a ≡ b mod n, then b ≡ a mod n 3. If a ≡ b mod n and b ≡ c mod n, then a ≡ c mod n 4. If a ≡ b mod n and c ≡ d mod n , then a+c ≡ b+d mod n and ac ≡ bd mod n 5. If a ≡ b mod n, then a+c ≡ b+c mod n and ac ≡ bc mod n.

Part 2.Views From Some ProblemsGeometric Constructions • The ancient Greeks were fond of geometric constructions. • They were especially interested in constructions that could be achieved using only a straightedge without markings and a compass. • Questions: 1. Can you bisect any given angel? 2. Can you trisect any given angel? 3. For any given positive integer n, can you construct a regular polygon of n sides? 4. Given a circle, is it possible to construct a square with same area? 5. Given a side of a cube, is it possible to construct a cube which has double the volume of the original cube?

Q1. Construction: Bisect Angle • Step 1. Draw an arc that is centered at the vertex of the angle. This arc can have a radius of any length. However, it must intersect both sides of the angle. We will call these intersection points P and Q This provides a point on each line that is an equal distance from the vertex of the angle. • Step 2. Draw two more arcs. The first arc must be centered on one of the two points P or Q. It can have any length radius. The second arc must be centered on whichever point (P or Q) you did NOT choose for the first arc. The radius for the second arc MUST be the same as the first arc. Make sure you make the arcs long enough so that these two arcs intersect in at least one point. We will call this intersection point X. Every intersection point between these arcs (there can be at most 2) will lie on the angle bisector. • Step 3. Draw a line that contains both the vertex and X. Since the intersection points and the vertex all lie on the angle bisector, we know that the line which passes through these points must be the angle bisector. • Now, try to do this construction yourself.

We can use field theory to prove it: • In 1801, Gauss asserted that a regular polygon of n sides is constructible if and only if n has the form 2k P1P2…Ps, where Pi’s are distinct primes of the form 2t +1. Example: regular polygons with 3,4,5,6,8,10,12,15,16,17 and 20 sides are constructible. But regular polygons with 7,9,11,13,14,18,19 sides are not. 2

One Example In Chemistry: • A benzene (苯) molecule can be modeled as six carbon arranged in a regular hexagon in a plane. At each carbon atom, one of three radicals NH2, COOH, or OH can be attached. How many such compounds are possible? • We can use Burnside’s theorem (one theorem coming from group action) to answer it. • There are also some interesting problems for coloring or counting.

Some Examples of Groups

Solvability of Polynomials by Radicals: • The equation • The formulas for general cubic and quartic (fourth-degree polynomial) are more complicated, but they can be given in terms of radiacls of rational expressions of the coefficients. • Is it solvable by radicals for a quintic ( fifth-degree polynomial)? Use Galois theory we can prove that 3x5 - 15x+5 is not solvable by radicals.

應用代數學 簡介