Math 412, Introduction to abstract algebra. Overview of algebra.

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1 Math 412, Introduction to abstract algebra. Overview of algebra. A study of algebraic objects and functions between them; an algebraic object is typically a set with one or more operations which satisfies certain axioms. Examples: vector space, eg.r 4 Main types of objects for this course: ring, eg. Z, the ring of integers having 2 operations; M n (R) in which multiplication is not commutative field, eg. Ror Q (a special type of ring; commutative and have multiplicative inverses) group, eg.z 3,S 3 ; 1 operation, commutative or not Groups have the simplest set of axioms, but are not necessarily the easiest to understand. Functions between these must preserve the operations: eg., linear transformations on vector spaces. We will have special names for the functions we wish to work with in each case (isomorphism, iso = equal, homomorphism, homo = same) To study such functions, we need to understand the kernels, special subsets which get mapped to zero (nullspaces for vector spaces, ideals for rings, normal subgroups). Chapter 2 of the book is about modular arithmetic, not because it is interesting in number theory, but because it is closely related to understanding these kernels. A topic that we will do only a little with (see grad algebra for more) is building new rings/groups out of simpler ones. For example, using Cartesian products ( 8.1 for groups). There are a number of standard methods of construction that apply to all types of algebraic objects this leads to the general idea of category theory to do them all at once. Algebraists love to generalize and find common properties in different objects. In fact, if there is one overlying theme in all of algebra, this is it: answering the question in what way are two different things the same. Historically (early in the 1900 s) abstract algebra had its beginnings with applications to algebraic geometry, algebraic topology and algebraic number theory, using algebraic ideas to help in understanding more traditional areas in which there were problems of interest to people. Thus abstract algebra played the same role in math that math has played in science, being a tool to organize the ideas. Where does high school algebra fit in? One generally studies polynomial expressions in high school. These are the elements of polynomial rings, such as R[x] ={ a i x i a i R} and will be studied in Chapters 4 and 5. This can also be generalized to polynomials in more than one variable, such as R[x, y]. As you go through this semester/year, you should be constantly asking yourselves, how 1

2 2 is what I am seeing now like something I have done before? You should always be trying to understand new definitions (of which there will be a lot) by thinking of examples you are familiar with, as well as why similar constructions are not examples. For example, Z is a commutative ring, but M n (R) isnot. Zhas a multiplicative cancellation law, but M n (R) does not. Chapter 1, Arithmetic in Z revisited. Theorem 1.1. (Division algorithm) Let a, b be integers with b>0. unique integers q, r such that Then there exist a = bq + r and 0 r<b Proof. Let S = { a bx x Z, a bx 0 } Note that S since x = a gives because b 1. a bx = a + b a = a (b ± 1) 0 Use the Well-ordering Axiom: every nonempty subset of the set of nonnegative integers contains a smallest element. (See Appendix C.) This says S has some smallest element r, say r= a bq, where qis the value of x giving r as an element of S. Wenowhavea=bq + r and r 0. We still need r<band uniqueness. Consider the integer a b(q + 1). If it is nonnegative, it lies in S; but a b(q +1)=a bq b = r b<r,

3 3 and r is the smallest element of S. Therefore, r b = a b(q +1)<0, so r<b. Uniqueness: Assume there are (possibly different) numbers q 1 and r 1 such that a = bq 1 + r 1 and 0 r 1 <b.then bq + r = a = bq 1 + r 1, so b(q q 1 )=r 1 r. This last equality can only hold for 0 r, r 1 <bif q = q 1 and r = r 1 (draw picture or see book s inequalities, p. 5). This is fundamental to the Euclidean Algorithm: Theorem 1.6. (Euclidean algorithm) Let a, b be positive integers with a b. If b a,then gcd(a, b) =b. Otherwise, apply the division algorithm repeatedly as follows; there is always an integer t such that r t is the last nonzero remainder in Then r t =gcd(a, b). a = bq 0 + r 0, 0 <r 0 <b b=r 0 q 1 +r 1, 0<r 1 <r 0 r 0 =r 1 q 2 +r 2, 0<r 2 <r 1 r t 2 =r t 1 q t +r t, r t 1 =r t q t+1 +0 To understand this, we need some definitions.. 0<r t <r t 1 Definition, p. 7. Given a, b Z, b 0,wesaybdivides a or b is a divisor or factor of a if there exists c Z such that a = bc. Notation b a. Note that, if a 0, b a,andsoahas only finitely many divisors. a =0, a =6. Examples: Definition, p. 8. common divisor; greatest common divisor (gcd). Write (a, b) or gcd(a, b). We say a and b are relatively prime if (a, b) =1. Examples: (4, 6) = 2, (5, 9) = 1. Sketch of proof of Theorem 1.6. The process stops in at most b steps since the r i s decrease. Check that r t is a common divisor and that any common divisor of a and b must divide r t.

4 4 A theorem which will be very useful later in studying ideals. Theorem 1.3. Let a, b Z, notboth0, and let d =gcd(a, b). Then there exist integers u, v with d = au + bv. Furthermore, d is the smallest positive integer that can be written in the form au + bv. Proof. The book gives a non-constructive proof, getting existence from the Well-ordering axiom. Another way is to use the Euclidean algorithm: d = r t = r t 2 r t 1 q t = r t 2 (r t 3 r t 2 q t 1 )q t = r t 2 (1 + q t 1 q t ) r t 3 = Replacing each of the r i s up to r 0 = a bq 0 gives an expression of the form au + bv where u and v are combinations of q i s. Now a = da 1 and b = db 1 for some a 1,b 1 so any positive integer of the form au + bv = d(a 1 u + b 1 v) d since a 1 u + b 1 v 1tomaketheau + bv positive. Example. Apply the Euclidean Algorithm to 114, 42 to get 6 = gcd(114, 42) = ( 8) 42. This is not unique: for example, we also have 6 = ( 4) Some standard number theory facts: Corollary 1.4 (the important part). If a, b Z are not both 0 and c a and c b, then c gcd(a, b). Proof. Write d =gcd(a, b) =au + bv = ca 1 u + cb 1 v = c(a 1 u + b 1 v). Theorem 1.5. If a bc and gcd(a, b) =1,then a c. Proof. By Theorem 1.3, we can write 1 = au + bv, so (using the hypothesis to write ar = bc), c = cau + cbv = cau + arv = a(cu + rv), and therefore a c.

5 5 Definition. An integer p is prime if p ±1 and the only divisors of p are ±1 and±p. Note that p 0. Comment on ±1 (units, p. 60). This will be generalized to arbitrary commutative rings in Chapter 6. (It can be done for noncommutative rings also.) If a number is not 0, ±1 and not prime, it is called composite. Theorem 1.8. Let p Z, p 0,±1.Thenpis prime iff p satisfies the property whenever p bc, thenp bor p c. Proof. (= ) Suppose p bc. Sincepis prime, its divisor gcd(p, b) mustbe1or p.ifitis p,then p b.if it is 1,then p cby Theorem 1.5. ( =) Let d be a divisor of p, sayp=ds. By hypothesis, p d or p s. Ifp s,says=ps 1, then p = ds = pds 1,sods 1 =1andd=±1. Otherwise, p d and d p, hence d = ±p. By definition, p is prime. Corollary 1.9. If p is prime and p a 1 a 2 a n,thenpdivides at least one of the a i. Proof. The book s proof is an informal one. To write it carefully, we need mathematical induction. The statement holds for n = 2 by Theorem 1.8. Assume it is true for n 1and n 3. Show it holds for n... Theorem 1.11 (Fundamental Theorem of Arithmetic). Every integer n 0,±1 can be written as a product of primes. The prime factorization is unique up to rearranging the factors and multiplication of the factors by ±1. Comment on proof by contradiction; i.e., use of the contrapositive to a statement. Proof. It clearly suffices to prove the theorem for n>1. Assume that some integer cannot be written as a product of primes. Let m be the smallest positive integer that cannot be so written. In particular, m is not prime, so it has positive factors other than 1 and m, saym=ab with 1 <a,b<m. Since a and b are positive integers less than m, they can be written as products of primes, say a = p 1 p r and b = q 1 q s. But then m = p 1 p r q 1 q s, a contradiction of the choice of m. Now assume that n has two prime factorizations n = p 1 p r =q 1 q s. Then p 1 divides q 1 q s, hence by Corollary 1.9, p 1 divides some q i ; reordering if necessary, we may assume i =1. Butq 1 is prime, so p 1 = ±q 1. Dividing both sides by q 1 gives us p 2 (±p 3 p r )=q 2 q s.

6 6 Use the same argument to conclude that (reordering if necessary) p 2 = ±q 2. Continue until all the primes are cancelled from one side. For example, if s>r, we end up with ±1 =q s r q s, which is impossible as the q i s are all prime. Thus we must have r = s and the factorization has the claimed uniqueness. Examples. From pages 18-19, do problems 23, 27 (generalizing part (b)), 12(a) (which leads to Chapter 2 work).